quadratic relations assignment

MPM2D: PRINCIPLES OF MATHEMATICS GRADE 10 , ACADEMIC SAMPLE EXERCISES ☰ Menu: Click to navigate this page

Linear systems.

• Graphical method
• Substitution method
• Elimination method
• Consistency vs inconsistency

Analytic geometry

• Midpoint of a line segment
• Length of a line segment
• Median of a triangle
• Right bisectors
• Equation of a circle

Properties of quadratic relations

• Graphing quadratics
• Transformations of the parent quadratic
• Sketching quadratics by hand
• Equation of a parabola

Applications of quadratic relations

• Expanding binomials
• Factoring quadratics
• Factors and x-intercepts
• Real and non-real roots
• Completing the square
• Solving quadratic equations
• Some word problems involving quadratics

Exercises on trigonometry

• ASSIGNMENT on trigonometry
• QUIZ on similar triangles
• Sample TEST on trigonometry
• Class exercise (#22) on applications of similar triangles
• Class exercise (#23) on applications of right triangles
• Class exercise (#24) on applications of oblique triangles

Sample Final Exam

• Sample final exam

Exercises on linear systems

• ASSIGNMENT on linear systems using elimination
• QUIZ on linear systems
• Sample TEST on linear systems
• Class exercise (#1) on graphical solution of linear systems
• Class exercise (#2) on substitution method
• Class exercise (#3) on elimination method
• Class exercise (#4) on some application of linear systems

Exercises on analytic geometry

• ASSIGNMENT on analytic geometry
• Another ASSIGNMENT on analytic geometry
• Sample TEST on analytic geometry
• Class exercise (#5) on length of a line segment
• Class exercise (#6) on midpoint of a line segment
• Class exercise (#7) on bisectors and medians
• Class exercise (#8) on some geometric properties
• Class exercise (#9) on equation of a circle

Exercises on properties of quadratic relations

• ASSIGNMENT on quadratic relations
• QUIZ on quadratic relations
• Sample TEST on quadratic relations
• Class exercise (#10) on graphing quadratics
• Class exercise (#11) on key features of a parabola
• Class exercise (#12) on transformations of the parent quadratic
• Class exercise (#13) on sketching parabolas by hand
• Class exercise (#14) on equation of a parabola

Exercises on applications of quadratic relations

• ASSIGNMENT on factoring quadratics
• Another ASSIGNMENT on quadratic equations and applications
• Sample TEST on quadratic equations and applications
• Class exercise (#15) on expanding binomials
• Class exercise (#16) on factoring quadratics
• Class exercise (#17) on connecting factors and x-intercepts
• Class exercise (#18) on real and non-real roots
• Class exercise (#19) on completing the square
• Class exercise (#20) on sketching quadratics
• Class exercise (#21) on solving quadratic equations

Maybe you've experienced this as a teacher (irrespective of your teaching subject). You designed a home work or an assignment. You printed copies and distributed to your students. You gave them a day to submit the work. The day arrived. Then a particular student -- or some students -- didn't do the work. You asked why? In response, the student mentioned that (inadvertently) the work was misplaced (or lost). The student then asked for another copy of the exercise. You, the teacher, obliged and gave another copy of the exercise to the student. The incidence then repeats itself. This is one of the excuses some students give for not doing the exercises assigned to them. From our personal perspective, not many classroom experiences frustrate as much as the above. In a bid to obviate this, we came up with this website -- our initial objective was to make all exercises available to our students outside the classroom (and school) environment. Our approach may have been a drastic one, but it was also made possible by our interest in web development. We then have to print our exercises ONCE and only ONCE. If the student misplaces (or loses) the copy of the exercise, it is then the student's responsibility to print new copies from our website. It will follow that most of the exercises below reflect our teaching taste, and so may not appeal to a wide audience. In fact, they are patterned after the Ontario curriculum, and so their possible usefulness may be limited to this Canadian province. Also, we haven't provided solutions to them -- but we're working on this, including changing the format. In the meantime if you find any of these exercises useful, we'll be more than excited.

5.1 Quadratic Functions

Learning objectives.

In this section, you will:

• Recognize characteristics of parabolas.
• Understand how the graph of a parabola is related to its quadratic function.
• Determine a quadratic function’s minimum or maximum value.
• Solve problems involving a quadratic function’s minimum or maximum value.

Curved antennas, such as the ones shown in Figure 1 , are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.

Recognizing Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola . One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure 2 .

The y -intercept is the point at which the parabola crosses the y -axis. The x -intercepts are the points at which the parabola crosses the x -axis. If they exist, the x -intercepts represent the zeros , or roots , of the quadratic function, the values of x x at which y = 0. y = 0.

Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y - y - intercept of the parabola shown in Figure 3 .

The vertex is the turning point of the graph. We can see that the vertex is at ( 3 , 1 ) . ( 3 , 1 ) . Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. x = 3. This parabola does not cross the x - x - axis, so it has no zeros. It crosses the y - y - axis at ( 0 , 7 ) ( 0 , 7 ) so this is the y -intercept.

Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions

The general form of a quadratic function presents the function in the form

where a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0. If a > 0 , a > 0 , the parabola opens upward. If a < 0 , a < 0 , the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by x = − b 2 a . x = − b 2 a . If we use the quadratic formula, x = − b ± b 2 − 4 a c 2 a , x = − b ± b 2 − 4 a c 2 a , to solve a x 2 + b x + c = 0 a x 2 + b x + c = 0 for the x - x - intercepts, or zeros, we find the value of x x halfway between them is always x = − b 2 a , x = − b 2 a , the equation for the axis of symmetry.

Figure 4 represents the graph of the quadratic function written in general form as y = x 2 + 4 x + 3. y = x 2 + 4 x + 3. In this form, a = 1 , b = 4 , a = 1 , b = 4 , and c = 3. c = 3. Because a > 0 , a > 0 , the parabola opens upward. The axis of symmetry is x = − 4 2 ( 1 ) = −2. x = − 4 2 ( 1 ) = −2. This also makes sense because we can see from the graph that the vertical line x = −2 x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, ( −2 , −1 ) . ( −2 , −1 ) . The x - x - intercepts, those points where the parabola crosses the x - x - axis, occur at ( −3 , 0 ) ( −3 , 0 ) and ( −1 , 0 ) . ( −1 , 0 ) .

The standard form of a quadratic function presents the function in the form

where ( h , k ) ( h , k ) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function .

As with the general form, if a > 0 , a > 0 , the parabola opens upward and the vertex is a minimum. If a < 0 , a < 0 , the parabola opens downward, and the vertex is a maximum. Figure 5 represents the graph of the quadratic function written in standard form as y = −3 ( x + 2 ) 2 + 4. y = −3 ( x + 2 ) 2 + 4. Since x – h = x + 2 x – h = x + 2 in this example, h = –2. h = –2. In this form, a = −3 , h = −2 , a = −3 , h = −2 , and k = 4. k = 4. Because a < 0 , a < 0 , the parabola opens downward. The vertex is at ( − 2 , 4 ) . ( − 2 , 4 ) .

The standard form is useful for determining how the graph is transformed from the graph of y = x 2 . y = x 2 . Figure 6 is the graph of this basic function.

If k > 0 , k > 0 , the graph shifts upward, whereas if k < 0 , k < 0 , the graph shifts downward. In Figure 5 , k > 0 , k > 0 , so the graph is shifted 4 units upward. If h > 0 , h > 0 , the graph shifts toward the right and if h < 0 , h < 0 , the graph shifts to the left. In Figure 5 , h < 0 , h < 0 , so the graph is shifted 2 units to the left. The magnitude of a a indicates the stretch of the graph. If | a | > 1 , | a | > 1 , the point associated with a particular x - x - value shifts farther from the x- axis, so the graph appears to become narrower, and there is a vertical stretch. But if | a | < 1 , | a | < 1 , the point associated with a particular x - x - value shifts closer to the x- axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5 , | a | > 1 , | a | > 1 , so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

For the linear terms to be equal, the coefficients must be equal.

This is the axis of symmetry we defined earlier. Setting the constant terms equal:

In practice, though, it is usually easier to remember that k is the output value of the function when the input is h , h , so f ( h ) = k . f ( h ) = k .

Forms of Quadratic Functions

A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.

The general form of a quadratic function is f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c where a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0.

The standard form of a quadratic function is f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k where a ≠ 0. a ≠ 0.

The vertex ( h , k ) ( h , k ) is located at

Given a graph of a quadratic function, write the equation of the function in general form.

• Identify the horizontal shift of the parabola; this value is h . h . Identify the vertical shift of the parabola; this value is k . k .
• Substitute the values of the horizontal and vertical shift for h h and k . k . in the function f ( x ) = a ( x – h ) 2 + k . f ( x ) = a ( x – h ) 2 + k .
• Substitute the values of any point, other than the vertex, on the graph of the parabola for x x and f ( x ) . f ( x ) .
• Solve for the stretch factor, | a | . | a | .
• Expand and simplify to write in general form.

Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function g g in Figure 7 as a transformation of f ( x ) = x 2 , f ( x ) = x 2 , and then expand the formula, and simplify terms to write the equation in general form.

We can see the graph of g is the graph of f ( x ) = x 2 f ( x ) = x 2 shifted to the left 2 and down 3, giving a formula in the form g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3. g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3.

Substituting the coordinates of a point on the curve, such as ( 0 , −1 ) , ( 0 , −1 ) , we can solve for the stretch factor.

In standard form, the algebraic model for this graph is ( g ) x = 1 2 ( x + 2 ) 2 – 3. ( g ) x = 1 2 ( x + 2 ) 2 – 3.

To write this in general polynomial form, we can expand the formula and simplify terms.

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

We can check our work using the table feature on a graphing utility. First enter Y1 = 1 2 ( x + 2 ) 2 − 3. Y1 = 1 2 ( x + 2 ) 2 − 3. Next, select TBLSET, TBLSET, then use TblStart = – 6 TblStart = – 6 and Δ Tbl = 2, Δ Tbl = 2, and select TABLE . TABLE . See Table 1 .

The ordered pairs in the table correspond to points on the graph.

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8 . Assume that the point (–4, 7) is the highest point of the basketball’s trajectory. Find an equation for the path of the ball. Does the shooter make the basket?

Given a quadratic function in general form, find the vertex of the parabola.

• Identify a ,   b ,   and   c . a ,   b ,   and   c .
• Find h , h , the x -coordinate of the vertex, by substituting a a and b b into h = – b 2 a . h = – b 2 a .
• Find k , k , the y -coordinate of the vertex, by evaluating k = f ( h ) = f ( − b 2 a ) . k = f ( h ) = f ( − b 2 a ) .

Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function f ( x ) = 2 x 2 – 6 x + 7. f ( x ) = 2 x 2 – 6 x + 7. Rewrite the quadratic in standard form (vertex form).

The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2 The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2

Rewriting into standard form, the stretch factor will be the same as the a a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the " a a " from the general form.

The standard form of a quadratic function prior to writing the function then becomes the following:

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k , k , and where it occurs, x . x .

Given the equation g ( x ) = 13 + x 2 − 6 x , g ( x ) = 13 + x 2 − 6 x , write the equation in general form and then in standard form.

Finding the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y -values greater than or equal to the y -coordinate at the turning point or less than or equal to the y -coordinate at the turning point, depending on whether the parabola opens up or down.

Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions.

The range of a quadratic function written in general form f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c with a positive a a value is f ( x ) ≥ f ( − b 2 a ) , f ( x ) ≥ f ( − b 2 a ) , or [ f ( − b 2 a ) , ∞ ) ; [ f ( − b 2 a ) , ∞ ) ; the range of a quadratic function written in general form with a negative a a value is f ( x ) ≤ f ( − b 2 a ) , f ( x ) ≤ f ( − b 2 a ) , or ( − ∞ , f ( − b 2 a ) ] . ( − ∞ , f ( − b 2 a ) ] .

The range of a quadratic function written in standard form f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k with a positive a a value is f ( x ) ≥ k ; f ( x ) ≥ k ; the range of a quadratic function written in standard form with a negative a a value is f ( x ) ≤ k . f ( x ) ≤ k .

Given a quadratic function, find the domain and range.

• Identify the domain of any quadratic function as all real numbers.
• Determine whether a a is positive or negative. If a a is positive, the parabola has a minimum. If a a is negative, the parabola has a maximum.
• Determine the maximum or minimum value of the parabola, k . k .
• If the parabola has a minimum, the range is given by f ( x ) ≥ k , f ( x ) ≥ k , or [ k , ∞ ) . [ k , ∞ ) . If the parabola has a maximum, the range is given by f ( x ) ≤ k , f ( x ) ≤ k , or ( − ∞ , k ] . ( − ∞ , k ] .

Find the domain and range of f ( x ) = − 5 x 2 + 9 x − 1. f ( x ) = − 5 x 2 + 9 x − 1.

As with any quadratic function, the domain is all real numbers.

Because a a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x - x - value of the vertex.

The maximum value is given by f ( h ) . f ( h ) .

The range is f ( x ) ≤ 61 20 , f ( x ) ≤ 61 20 , or ( − ∞ , 61 20 ] . ( − ∞ , 61 20 ] .

Find the domain and range of f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 . f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 .

Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola . We can see the maximum and minimum values in Figure 9 .

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

• ⓐ Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L . L .
• ⓑ What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W , W , to represent the width of the garden and the length of the fence section parallel to the backyard fence.

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

This formula represents the area of the fence in terms of the variable length L . L . The function, written in general form, is

• ⓑ The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a a is the coefficient of the squared term, a = −2 , b = 80 , a = −2 , b = 80 , and c = 0. c = 0.

To find the vertex:

The maximum value of the function is an area of 800 square feet, which occurs when L = 20 L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11 .

Given an application involving revenue, use a quadratic equation to find the maximum.

• Write a quadratic equation for a revenue function.
• Find the vertex of the quadratic equation.
• Determine the y -value of the vertex.

Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p p for price per subscription and Q Q for quantity, giving us the equation Revenue = p Q . Revenue = p Q .

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 p = 30 and Q = 84,000. Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 p = 32 and Q = 79,000. Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y -intercept.

This gives us the linear equation Q = −2,500 p + 159,000 Q = −2,500 p + 159,000 relating cost and subscribers. We now return to our revenue equation.

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

This could also be solved by graphing the quadratic as in Figure 12 . We can see the maximum revenue on a graph of the quadratic function.

Finding the x - and y -Intercepts of a Quadratic Function

Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y - y - intercept of a quadratic by evaluating the function at an input of zero, and we find the x - x - intercepts at locations where the output is zero. Notice in Figure 13 that the number of x - x - intercepts can vary depending upon the location of the graph.

Given a quadratic function f ( x ) , f ( x ) , find the y - y - and x -intercepts.

• Evaluate f ( 0 ) f ( 0 ) to find the y -intercept.
• Solve the quadratic equation f ( x ) = 0 f ( x ) = 0 to find the x -intercepts.

Finding the y - and x -Intercepts of a Parabola

Find the y - and x -intercepts of the quadratic f ( x ) = 3 x 2 + 5 x − 2. f ( x ) = 3 x 2 + 5 x − 2.

We find the y -intercept by evaluating f ( 0 ) . f ( 0 ) .

So the y -intercept is at ( 0 , −2 ) . ( 0 , −2 ) .

For the x -intercepts, we find all solutions of f ( x ) = 0. f ( x ) = 0.

In this case, the quadratic can be factored easily, providing the simplest method for solution.

So the x -intercepts are at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( − 2 , 0 ) . ( − 2 , 0 ) .

By graphing the function, we can confirm that the graph crosses the y -axis at ( 0 , −2 ) . ( 0 , −2 ) . We can also confirm that the graph crosses the x -axis at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( −2 , 0 ) . ( −2 , 0 ) . See Figure 14

Rewriting Quadratics in Standard Form

In Example 7 , the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

Given a quadratic function, find the x - x - intercepts by rewriting in standard form .

• Substitute a a and b b into h = − b 2 a . h = − b 2 a .
• Substitute x = h x = h into the general form of the quadratic function to find k . k .
• Rewrite the quadratic in standard form using h h and k . k .
• Solve for when the output of the function will be zero to find the x - x - intercepts.

Finding the x -Intercepts of a Parabola

Find the x - x - intercepts of the quadratic function f ( x ) = 2 x 2 + 4 x − 4. f ( x ) = 2 x 2 + 4 x − 4.

We begin by solving for when the output will be zero.

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

We know that a = 2. a = 2. Then we solve for h h and k . k .

So now we can rewrite in standard form.

We can now solve for when the output will be zero.

The graph has x -intercepts at ( −1 − 3 , 0 ) ( −1 − 3 , 0 ) and ( −1 + 3 , 0 ) . ( −1 + 3 , 0 ) .

We can check our work by graphing the given function on a graphing utility and observing the x - x - intercepts. See Figure 15 .

We could have achieved the same results using the quadratic formula. Identify a = 2 , b = 4 a = 2 , b = 4 and c = −4. c = −4.

So the x -intercepts occur at ( − 1 − 3 , 0 ) ( − 1 − 3 , 0 ) and ( − 1 + 3 , 0 ) . ( − 1 + 3 , 0 ) .

In a Try It , we found the standard and general form for the function g ( x ) = 13 + x 2 − 6 x . g ( x ) = 13 + x 2 − 6 x . Now find the y - and x -intercepts (if any).

Applying the Vertex and x -Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H ( t ) = − 16 t 2 + 80 t + 40. H ( t ) = − 16 t 2 + 80 t + 40.

• ⓐ When does the ball reach the maximum height?
• ⓑ What is the maximum height of the ball?
• ⓒ When does the ball hit the ground?

The ball reaches a maximum height after 2.5 seconds.

The ball reaches a maximum height of 140 feet.

We use the quadratic formula.

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16 .

Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H ( t ) = −16 t 2 + 96 t + 112. H ( t ) = −16 t 2 + 96 t + 112.

• ⓐ When does the rock reach the maximum height?
• ⓑ What is the maximum height of the rock?
• ⓒ When does the rock hit the ocean?

Access these online resources for additional instruction and practice with quadratic equations.

• Graphing Quadratic Functions in General Form
• Graphing Quadratic Functions in Standard Form
• Quadratic Function Review
• Characteristics of a Quadratic Function

5.1 Section Exercises

Explain the advantage of writing a quadratic function in standard form.

How can the vertex of a parabola be used in solving real-world problems?

Explain why the condition of a ≠ 0 a ≠ 0 is imposed in the definition of the quadratic function.

What is another name for the standard form of a quadratic function?

What two algebraic methods can be used to find the horizontal intercepts of a quadratic function?

For the following exercises, rewrite the quadratic functions in standard form and give the vertex.

f ( x ) = x 2 − 12 x + 32 f ( x ) = x 2 − 12 x + 32

g ( x ) = x 2 + 2 x − 3 g ( x ) = x 2 + 2 x − 3

f ( x ) = x 2 − x f ( x ) = x 2 − x

f ( x ) = x 2 + 5 x − 2 f ( x ) = x 2 + 5 x − 2

h ( x ) = 2 x 2 + 8 x − 10 h ( x ) = 2 x 2 + 8 x − 10

k ( x ) = 3 x 2 − 6 x − 9 k ( x ) = 3 x 2 − 6 x − 9

f ( x ) = 2 x 2 − 6 x f ( x ) = 2 x 2 − 6 x

f ( x ) = 3 x 2 − 5 x − 1 f ( x ) = 3 x 2 − 5 x − 1

For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry.

y ( x ) = 2 x 2 + 10 x + 12 y ( x ) = 2 x 2 + 10 x + 12

f ( x ) = 2 x 2 − 10 x + 4 f ( x ) = 2 x 2 − 10 x + 4

f ( x ) = − x 2 + 4 x + 3 f ( x ) = − x 2 + 4 x + 3

f ( x ) = 4 x 2 + x − 1 f ( x ) = 4 x 2 + x − 1

h ( t ) = −4 t 2 + 6 t − 1 h ( t ) = −4 t 2 + 6 t − 1

f ( x ) = 1 2 x 2 + 3 x + 1 f ( x ) = 1 2 x 2 + 3 x + 1

f ( x ) = − 1 3 x 2 − 2 x + 3 f ( x ) = − 1 3 x 2 − 2 x + 3

For the following exercises, determine the domain and range of the quadratic function.

f ( x ) = ( x − 3 ) 2 + 2 f ( x ) = ( x − 3 ) 2 + 2

f ( x ) = −2 ( x + 3 ) 2 − 6 f ( x ) = −2 ( x + 3 ) 2 − 6

f ( x ) = x 2 + 6 x + 4 f ( x ) = x 2 + 6 x + 4

f ( x ) = 2 x 2 − 4 x + 2 f ( x ) = 2 x 2 − 4 x + 2

For the following exercises, use the vertex ( h , k ) ( h , k ) and a point on the graph ( x , y ) ( x , y ) to find the general form of the equation of the quadratic function.

( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 ) ( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 )

( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 ) ( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 )

( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 )

( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 ) ( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 )

( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 ) ( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 )

( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 ) ( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 )

( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 )

( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 ) ( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 )

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.

f ( x ) = x 2 − 2 x f ( x ) = x 2 − 2 x

f ( x ) = x 2 − 6 x − 1 f ( x ) = x 2 − 6 x − 1

f ( x ) = x 2 − 5 x − 6 f ( x ) = x 2 − 5 x − 6

f ( x ) = x 2 − 7 x + 3 f ( x ) = x 2 − 7 x + 3

f ( x ) = −2 x 2 + 5 x − 8 f ( x ) = −2 x 2 + 5 x − 8

f ( x ) = 4 x 2 − 12 x − 3 f ( x ) = 4 x 2 − 12 x − 3

For the following exercises, write the equation for the graphed quadratic function.

For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.

For the following exercises, use a calculator to find the answer.

Graph on the same set of axes the functions f ( x ) = x 2 f ( x ) = x 2 , f ( x ) = 2 x 2 f ( x ) = 2 x 2 , and f ( x ) = 1 3 x 2 f ( x ) = 1 3 x 2 .

What appears to be the effect of changing the coefficient?

Graph on the same set of axes f ( x ) = x 2 , f ( x ) = x 2 + 2 f ( x ) = x 2 , f ( x ) = x 2 + 2 and f ( x ) = x 2 , f ( x ) = x 2 + 5 f ( x ) = x 2 , f ( x ) = x 2 + 5 and f ( x ) = x 2 − 3. f ( x ) = x 2 − 3. What appears to be the effect of adding a constant?

Graph on the same set of axes f ( x ) = x 2 , f ( x ) = ( x − 2 ) 2 , f ( x − 3 ) 2 f ( x ) = x 2 , f ( x ) = ( x − 2 ) 2 , f ( x − 3 ) 2 , and f ( x ) = ( x + 4 ) 2 . f ( x ) = ( x + 4 ) 2 .

What appears to be the effect of adding or subtracting those numbers?

The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function h ( x ) = − 32 ( 80 ) 2 x 2 + x h ( x ) = − 32 ( 80 ) 2 x 2 + x where x x is the horizontal distance traveled and h ( x ) h ( x ) is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.

A suspension bridge can be modeled by the quadratic function h ( x ) = .0001 x 2 h ( x ) = .0001 x 2 with −2000 ≤ x ≤ 2000 −2000 ≤ x ≤ 2000 where | x | | x | is the number of feet from the center and h ( x ) h ( x ) is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet.

For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.

Vertex ( 1 , −2 ) , ( 1 , −2 ) , opens up.

Vertex ( −1 , 2 ) ( −1 , 2 ) opens down.

Vertex ( −5 , 11 ) , ( −5 , 11 ) , opens down.

Vertex ( −100 , 100 ) , ( −100 , 100 ) , opens up.

For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function.

Contains ( 1 , 1 ) ( 1 , 1 ) and has shape of f ( x ) = 2 x 2 . f ( x ) = 2 x 2 . Vertex is on the y - y - axis.

Contains ( −1 , 4 ) ( −1 , 4 ) and has the shape of f ( x ) = 2 x 2 . f ( x ) = 2 x 2 . Vertex is on the y - y - axis.

Contains ( 2 , 3 ) ( 2 , 3 ) and has the shape of f ( x ) = 3 x 2 . f ( x ) = 3 x 2 . Vertex is on the y - y - axis.

Contains ( 1 , −3 ) ( 1 , −3 ) and has the shape of f ( x ) = − x 2 . f ( x ) = − x 2 . Vertex is on the y - y - axis.

Contains ( 4 , 3 ) ( 4 , 3 ) and has the shape of f ( x ) = 5 x 2 . f ( x ) = 5 x 2 . Vertex is on the y - y - axis.

Contains ( 1 , −6 ) ( 1 , −6 ) has the shape of f ( x ) = 3 x 2 . f ( x ) = 3 x 2 . Vertex has x-coordinate of −1. −1.

Real-World Applications

Find the dimensions of the rectangular dog park producing the greatest enclosed area given 200 feet of fencing.

Find the dimensions of the rectangular dog park split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing.

Find the dimensions of the rectangular dog park producing the greatest enclosed area split into 3 sections of the same size given 500 feet of fencing.

Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product?

Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product?

Suppose that the price per unit in dollars of a cell phone production is modeled by p = $ 45 − 0.0125 x , p = $ 45 − 0.0125 x , where x x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ⋅ p . R = x ⋅ p . Find the production level that will maximize revenue.

A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by h ( t ) = −4.9 t 2 + 229 t + 234. h ( t ) = −4.9 t 2 + 229 t + 234. Find the maximum height the rocket attains.

A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h ( t ) = − 4.9 t 2 + 24 t + 8. h ( t ) = − 4.9 t 2 + 24 t + 8. How long does it take to reach maximum height?

A soccer stadium holds 62,000 spectators. With a ticket price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?

A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest?

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• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
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• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/5-1-quadratic-functions

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Quadratic relation, parabolas : translations and applications

PARABOLAS: TRANSLATIONS AND APPLICATIONS

QUADRATIC RELATION   A quadratic relation in two variables is a relation that can be written in the form

                y=ax^2+bx+c or  x=ay^2+by+c

where a , b , and c are real numbers, and a!=0 . The graphs of quadratic relations are called parabolas. The simplest quadratic relation of the form y=ax^2+bx+c is y=x^2 , with a=1 , b=0 , and c=0 , so this relation is graphed first.

Example 1 GRAPHING THE SIMPLEST QUADRATIC RELATION

Graph y = x^2    Set x equal to 0 in y = x^2 to get  y = 0 , which shows that the only intercept is at the origin. By plotting  other selected points, as shown in the table accompanying Figure 3.16, we obtain the graph. The domain is (-inf,inf) and the range is  [0,inf ]

    Notice in Figure 3.16 that the part of the graph in quadrant II is a “mirror image” of the part in quadrant I. We say that this graph is symmetric with respect to the y -axis. (More will be said about symmetry in the next section.) The line of symmetry for a parabola is the axis of the parabola. The lowest point on this parabola, the point (0,0) , is called the vertex of the parabola.

Starting with y = x^2 , there are several possible ways to get a more general expression:                      y = ax^2                Multiply by a positive or negative coefficient.                     y = x^2+k                positive or negative constant                      y = (x-h)^2          Replace  x with x-h , where his a constant                     y = a(x-h)^2+k            Do all of the above. The graph of each of these relations is still a parabola, but it is modified from that of y = x^2 . The next few examples show how these changes modify the parabola. The first example shows the result of changing y = x^2 to y = ax^2 .

Example 2 GRAPHING RELATIONS OF THE FORM y = ax^2

Graph each relation.

(a)  y = 2x^2

A table of selected ordered pairs is given with the graph in Figure 3.17. The y -values of the ordered pairs of this relation are twice as large as the corresponding y -values for the graph of y = x^2 . This makes the graph rise more rapidly, so the parabola is narrower than the parabola for y = x^2 , as can be seen in the figure.3.17

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(b)  y = -1/2x^2 The coefficient -1/2 causes the y -values to be closer to the x -axis than for y = x^2 , making the graph broader than that of y = x^2 . Because the y -values are negative for each nonzero x -value, this graph opens downward. Again the axis is the line x=0 , and the vertex, the highest point on the graph, is (0,0) . The domain is (-inf,inf) and the range is (-inf,0) . See Figure 3.18. 

As Example 2 suggests, |a|  in y = ax^2 determines the width of a parabola, so that it is narrower than the graph of y = x^2 . when  |a|>1   and broader than the graph of y = x^2 when |a|<1 .

The next two examples show how changing y = x^2 to y = x^2+k  or to y = (x-h)^2 , respectively, affects the graph of a parabola. Example 3 

GRAPHING A RELATION OF THE  y = x^2+k

Graph y = x^2-4  Each value of y will be 4 less than the corresponding value of y = x^2 . This means that y = x^2-4 has the same shape as y = x^2  but is shifted 4 units down. See Figure 3.19. The vertex of the parabola (on this parabola, the lowest point) is at (0,-4) . The axis of the parabola is the vertical line x=0 . When the vertex of a parabola is shifted vertically, the intercepts are usually good choices for additional points to plot. Here,

the y-intercept is -4 , which is the y -value of the vertex. The x -intercepts are found by setting y=0 :

                                  y=x^2-4

                                  0=x^2-4

                                  4=x^2

                          x=2 or  x=-2 The x -intercepts are 2 and -2 .

The vertical shift of the graph in Example 3 is called a translation. Example 4 below shows a horizontal translation, which is a shift to the right or left.

Example 4 GRAPHING A RELATION OF THE FORM  y={x-h}^2

Graph y=(x-4)^2  Comparing the two tables of ordered pairs shown in Figure 3.20, for y = x^2 and y=(x-4)^2 , indicates that this parabola is translated 4 units to the right as compared to the graph of y = x^2 . For example, the vertex is (4, 0)  instead of (0,0) , and x=-2 in y = x^2 corresponds to the same y -value as x=2 in y=(x-4)^2 , a difference of 2-(-2)=4 units. The axis of y=(x-4)^2 is the vertical line x=4 . See Figure 3.20.

A combination of all the transformations illustrated in Examples 2, 3, and 4 is shown in the next example.

Example 5 

GRAPHING A RELATION OF THE FORM  y=a(x-h)^2+k

Graph y=-(x+3)^2+1 .

This parabola is translated 3 units to the left and 1 unit up. Because of the negative sign, it opens downward, so that the vertex, the point (-3,1) , is the highest point on the graph. The axis is the line x=-3 . The y -intercept is  y=-(0+3)^2+1=-8 . By symmetry about the axis x=-3 , the point (-6,-8)  also is on the graph. The x -intercepts are found by solving the equation

                          0=-(x+3)^2+1

                          0=-(x^2+6x+9)+1       Square the binomial.                            0=-x^2-6x-9+1      Distributive property                           0=-x^2-6x-8                           0=x^2+6x+8            Multiply by —1.                           0=(x+2)(x+4) .       Factor. from which x=-2 or x=-4 . The graph is shown tn Figure 3.21. Noticed from the graph that the domain is (-inf,inf) and the range is (-inf,1) . The y -value of the vertex determines the range.

Examples 2-5 suggest the following generalizations.

GRAPH OF A PARABOLA

The graph of y=a(x-h)^2+k , where a!=0 , (a) is a parabola with vertex (h, k) , and the vertical line x=h as axis; (b) opens upward if a > 0   and downward if a < 0 ; (c) is broader than y = x^2 if 0 < |a| < 1  and narrower than y = x^2 if |a| > 1 . Given the relation y=ax^2+bx+c , where a , b , and c are real numbers and a!=0 , the process of completing the square (discussed in Chapter 2) can be used to change ax^2+bx+c to the form a(x-h)^2+k so that the vertex and axis may be identified. Follow the steps given in the next example.

Example 6 

GRAPHING PARABOLA BY COMPLETING THE SQUARE

Graph y=-3x^2-2x+1 .  A Our goal is to write the equation in the form y=a(x-h)^2+k . We may start by dividing both sides by -3 to get    y/-3=x^2+2/3x-1/3 Now complete the procedure, as explained in Section 2.4.                 y/-3+1/3=x^2+2/3x       Add  1/3 to both sides. 

    y/-3+1/3+1/9=x^2+2/3x+1/9   [1/2(2/3)]^2=1/9 , so add 1/9 to both side.                    y/-3+4/9=(x+1/3)^2               Combine terms on the left, Factor on the right.

   y/-3=(x+1/3)^2-4/9   Subtract  4/9 .

   y=-3(x+1/3)^2+4/3   Multiply by  -3 .

or   y=-3[x-(-1/3)]^2+4/3

   Figure 3.22

  Now the equation of the parabola is written in the form y = a(x - h)^2 + k , and this rewritten equation shows that the axis of the parabola is the vertical line x=-1/3 and that the vertex is (-1/3,4/3) . Use these results, together with the intercepts and additional ordered pairs as needed, to get the graph in Figure 3.22. From the graph, the domain of the relation is (-∞,∞) and the range is (-∞,4/3) .

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  A formula for the vertex of the graph of the quadratic relation y = ax + bx + c  can be found by completing the square for the general form of the equation.

   y=ax^2+bx+c    (a!=0)

   y/a=x^2+b/(a)x+c/a   Divide by  a

   y/a-c/a=x^2+b/(a)x   Subtract  c/a .

   y/a-c/a+(b^2)/(4a^2)=x^2+b/(a)x+(b^2)/(4a^2)   Add  (b^2)/(4a^2)

   y/a+(b^2-4ac)/(4a^2)=(x+b/(2a))^2   Combine terms on left and factor on right.

   y/a=(x+b/(2a))^2-(b^2-4ac)/(4a^2)   Get y term alone on the left.

   y=a(x+b/(2a))^2+(4ac-b^2)/(4a)   Multiply by  a

  The final equation shows that the vertex (h, k) can be expressed in terms of a,b , and c . However, it is not necessary to memorize the expression for k . since it can be obtained by replacing x with -b/(2a) .

   VERTEX OF A PARABOLA  (Y=ax^2+bx+c)

  The x -value of the vertex of the parabola y = ax^2 + bx + c , where a != 0 , is -b/(2a) .

USING THE VERTEX FORMULA

  Use the formula above to find the vertex of Lhe parabola y=2x^2-4x+3 .

  In this equation, a = 2 , b = -4 , and c = 3. By the formula given above, the x -value of the vertex of the parabola is 

   x=-b/(2a)=(-4)/(2(2))=1

  The y-value is found by substituting  1 for x into the equation y=2x^2-4x+3 to get y = 2(1)^2 - 4(1) + 3 = 1 , so the vertex is (1, 1) . 

  APPLICATION OF QUADRATIC RELATIONS Quadratic relations can be applied in situations as illustrated in the next example.

  PROBLEM SOLVING

  The fact that the vertex of a vertical parabola is the highest or lowest point on the graph makes equations of the form y = ax^2 + bx + c important in problems where the maximum or minimum value of some quantity is to be found. When a < 0 , the y -value of the vertex gives the maximum value of y and the x -value tells where it occurs. Similarly, when a>0 , the y -value of the vertex gives the minimum y -value. 

FINDING THE VERTEX IN AN APPLICATION

  Ms. Whitney owns and operates Aunt Emma‘s Pie Shop. She has hired a consultant to analyze her business operations. The consultant tells her that her profit P in dollars is given by

   P=120x-x^2

  where x is the number of units of pies that she makes. How many units of pies should be made in order lo maximize the profit? What is the maximum possible profit?

  The profit relation P can be rewritten as

   P=-x^2+120x+0 ,

  with a=-1 , b=120 , and c = 0 . The graph of this relation will be a parabola opening downward, so that the vertex, of the form (x, P) . will be the highest point on the graph. To find the vertex, use the fact that x = -b/(2a) :

   x=-120/(2(-1))=60

  Let x = 60 in the equation to find the value of P at the vertex.

   P=120(60)-60^2

   P=3600

  The vertex is (60, 3600) . Figure 3.23 shows the portion of the profit graph located in quadrant {Iota} . (Why is quadrant {Iota} the only one of interest here?) The maximum profit of $ 3600 occurs when 60 units of pies are made. In this case, profit increases as more and more pies are made up to 60 units and then decreases as more pies are made past this point. 

   Figure 3.23

  In this section, we started by defining a quadratic relation y = ax^2 + bx + c and, by point-plotting, found the graph, which we called a parabola. It is possible to start with a parabola. a set of points in the plane, and find the corresponding relation, using the formal geometric definition of a parabola.

   GEOMETRIC DEFINITION OF A PARABOLA Geometrically, a parabola is defined as the set of all points in a plane that are equally distant from a fixed point and a fixed line not containing the point. The point is called the focus and the line is the directrix. The line through the focus and perpendicular to the directrix is the axis of the parabola. The point on the axis that is equally distant from the focus and the directrix is the vertex of the parabola.

   Figure 3.24

  The parabola in Figure 3.24 has the point (0, p) as focus and the line y = -p  as directrix. The vertex is (0,0) . Let (x, y) be any point on the parabola. The distance from (x, y) to the directrix is |y - (-p)| , while the distance from (x, y) to (0, p) is root((x-0)^2+(y-p)^2) . Since (x, y) is equally distant from the directrix and the focus,

   |y - (-p)| = root((x-0)^2+(y-p)^2)

  Square both sides, getting

   (y+p)^2=x^2+(y-p)^2

   y^2+2py+p^2=x^2+y^2-2py+p^2

   4py=x^2

  the equation of the parabola with focus (0, p) and directrix y = -p . Solving 4py = x^2  for y gives

   y=1/(4p)x^2 ,

   Figure 3.25

   so that 1/(4p) = a when the equation is written in the form  y = ax^2+bx+c .

  This result could be extended to a parabola with vertex at (h, k) , focus p units above (h, k) , and directrix p units below (h, k) , or to a parabola with vertex at (h, k) , focus p units below (h, k) , and directrix p units above (h, k) .

  The geometric properties of parabolas lead to many practical applications. For example, if a light source is placed at the focus of a parabolic reflector, as in Figure 3.25, light rays reflect parallel to the axis, making a spotlight or flashlight. The process also works in reverse. Light rays from a distant source come in parallel to the axis and are reflected to a point at the focus. (If such a reflector is aimed at the sun, a temperature of several thousand degrees may be obtained.) This use of parabolic reflection is seen in the satellite dishes used to pick up signals from communications satellites.

   HORIZONTAL PARABOLAS The directrix of a parabola could be the vertical line x = -p , where p > 0 . with focus on the x -axis at (p, 0) , producing a parabola opening to the right. This parabola is the graph of the relation y^2 = 4px or x = [l/(4p)]y^ 2 . The next examples show the graphs of horizontal parabolas with equations of the form x = ay^2 + by + c .

GRAPHING A HORIZONTAL PARABOLA

   Figure 2.6

  Graph  x=y^2 .

  The equation x=y^2 can be obtained from y=x^2  by exchanging x and y . Choosing values of y and finding the corresponding values of x gives the parabola in Figure 3.26. The graph of x=y^2 , shown in red, is symmetric with respect to the line y = 0 and has vertex at (0,0) . For comparison. the graph of y=x^2 is shown in blue. These graphs are mirror images of each other with respect to the line y = x . From the graph, the domain of x=y^2 is (0,∞) , and the range is (-∞,∞) .

The domain and the range of a horizontal parabola, such as x = y^2 in Figure 3.26, can be determined by looking at the graph. Since the vertex (0,0) has the smallest x x-value of any point on the graph, and the graph extends indefinitely to the right. the domain is (0,∞) . Because the graph extends upward and downward indefinitely, the range is (-∞,∞) .

EXAMPLE 10 COMPLETING THE SQUARE TO GRAPH A HORIZONTAL PARABOLA   Graph  x=2y^2+6y+5 .

  To write this equation in the form x=a(y-k)^2+h , complete the square on y  as follows:

   x=2y^2+6y+5

   x/2-y^2+3y+5/2   Divide by  2

   x/2-5/2=y^2+3y   Subtract  5/2 .

   x/2-5/2+9/4=y^2+3y+9/4   Add  9/4 .

   x/2-1/4=(y+3/2)^2   Combine terms; factor.

   x/2=(y+3/2)^2+1/4   Add  1/4 .

   x=2(y+3/2)^2+1/2   Multiply by  2 .

  As this result shows. the vertex of the parabola is the point (1/2, -3/2) . The axis is the horizontal line

   y+3/2=0   or   y=-3/2 .

   Figure 2.7

  There is no y-intercept, since the vertex is on the right of the y -axis and the graph opens to the right. However, the x -intercept is

   x=2(0)^2+6(0)+5=5 .

  Using the vertex. the axis of symmetry, and the x -intercept, and plotting a few additional points gives the graph in Figure 3.27. The domain is (1/2, ∞) and the range is (-∞,∞) .

  The vertex of a horizontal parabola can also be found by using the values of a and b in x=ay^2 +by +0 .

VERTEX OF A PARABOLA  (x=ay^2+by+c)

  The y -value of the vertex of the parabola

   x=ay^2+by+c , where  a!=0 ,

  is  -b/(2a) The x -value is found by substitution of  -b/(2a) for  y .

CAUTION   Be careful when using the two vertex formulas of this section. It is essential that you recognize whether the parabola is a vertical parabola or a horizontal one, so that you can decide whether -b/(2a) represents the x - or y -coordinate of the vertex. (It always represents the coordinate of the variable that is squared.)

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Grades 9/10/11 Quadratic Relations

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Graphs and tables. Standard, factored, and vertex forms. Algebra of quadratic relations. Quadratic equations. Intersections of lines and parabolas.

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Unit 1: Basic Properties of Quadratic Relations

Unit 2: algebraic representations of quadratic relations, unit 3: algebraic skills, unit 4: graphing quadratic relations, unit 5: solving problems involving quadratic relations.

This lesson introduces quadratic relations to students by examining tables of values and second differences. 

This lesson extends our study of quadratic relations, with an emphasis on using second differences to determine unknown values in a table.

This lesson defines key terms related to parabolas (vertex, zeros, axis of symmetry, etc.) and interprets those terms within a given context.

Students will compare the features of the graphs of ‌\(y=x^2\) and ‌\(y=2^x\).

All of the pencil and paper practice exercises, answers, and solutions for this unit are reproduced here.

This is a collection of additional, and sometimes challenging, problems that extend the material covered in this unit, connect material from different lessons, and further explore real-world applications.

© 2018 University of Waterloo. Except where noted, all rights reserved.

Mrs. Smith's Class

“it is not that i'm so smart. but i stay with the questions much longer.” ~ albert einstein, unit 5 – characteristics of quadratic relations.

Tues. November 25 Lesson: TEST REVIEW for TEST TOMORROW! In-Class Assignment Review Questions: ASSIGNMENT REVIEW – Characteristics of Quadratics ; ASSIGNMENT REVIEW – Characteristics of Quadratics (SOLUTIONS)

Mon. November 24 Lesson:  Identifying Quadratics (Lesson Notes) Homework: Identifying Quadratics (PRACTICE QUESTIONS) ;  Identifying Quadratics (SOLUTIONS) In-Class Assignment Review Questions: ASSIGNMENT REVIEW – Characteristics of Quadratics ; ASSIGNMENT REVIEW – Characteristics of Quadratics (SOLUTIONS)

Fri. November 21 Lesson:  Word Problems (Lesson Notes) Homework: Quadratics Word Problems (PRACTICE QUESTIONS) ;  Word Problems (SOLUTIONS)

Thurs. November 20 Lesson:  Standard Form Continued (Lesson Notes) Homework: Standard Form Continued…(PRACTICE QUESTIONS) ;  Standard Form Continued…(SOLUTIONS)

Wed. November 19 Lesson:  Standard & Factored Form (Lesson Notes) Homework: Standard & Factored Form (PRACTICE QUESTIONS) ;  Standard & Factored Form (SOLUTIONS)

Tues. November 18 Lesson:  Vertex Form (Lesson Notes) Homework: Use Desmos Calculator to complete the Practice Questions – Vertex Form (PRACTICE QUESTIONS) ;  Vertex Form (SOLUTIONS)

Mon. November 17 Lesson: Key Features of Quadratics (Lesson Notes) Homework: Key Features of Quadratics (PRACTICE QUESTIONS) ;  Key Features of Quadratics (SOLUTIONS)

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7.7: Modeling with Quadratic Functions

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Learning Objectives

• Recognize characteristics of parabolas
• Understand how the graph of a parabola is related to its quadratic function
• Solve problems involving a quadratic function’s minimum or maximum value

in order to apply mathematical modeling to solve real-world applications.

Curved antennas, such as the ones shown in Figure \(\PageIndex{1}\), are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.

Recognizing Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure \(\PageIndex{2}\).

The y-intercept is the point at which the parabola crosses the \(y\)-axis. The x-intercepts are the points at which the parabola crosses the \(x\)-axis. If they exist, the x-intercepts represent the zeros , or roots , of the quadratic function, the values of \(x\) at which \(y=0\).

Example \(\PageIndex{1}\): Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in Figure \(\PageIndex{3}\).

The vertex is the turning point of the graph. We can see that the vertex is at \((3,1)\). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is \(x=3\). This parabola does not cross the x-axis, so it has no zeros. It crosses the \(y\)-axis at \((0,7)\) so this is the y-intercept.

Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions

The general form of a quadratic function presents the function in the form

\[f(x)=ax^2+bx+c\]

where \(a\), \(b\), and \(c\) are real numbers and \(a{\neq}0\). If \(a>0\), the parabola opens upward. If \(a<0\), the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by \(x=−\frac{b}{2a}\). If we use the quadratic formula, \(x=\frac{−b{\pm}\sqrt{b^2−4ac}}{2a}\), to solve \(ax^2+bx+c=0\) for the x-intercepts, or zeros, we find the value of \(x\) halfway between them is always \(x=−\frac{b}{2a}\), the equation for the axis of symmetry.

Figure \(\PageIndex{4}\) represents the graph of the quadratic function written in general form as \(y=x^2+4x+3\). In this form, \(a=1\), \(b=4\), and \(c=3\). Because \(a>0\), the parabola opens upward. The axis of symmetry is \(x=−\frac{4}{2(1)}=−2\). This also makes sense because we can see from the graph that the vertical line \(x=−2\) divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, \((−2,−1)\). The x-intercepts, those points where the parabola crosses the x-axis, occur at \((−3,0)\) and \((−1,0)\).

The standard form of a quadratic function presents the function in the form

\[f(x)=a(x−h)^2+k\]

where \((h, k)\) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function .

As with the general form, if \(a>0\), the parabola opens upward and the vertex is a minimum. If \(a<0\), the parabola opens downward, and the vertex is a maximum. Figure \(\PageIndex{5}\) represents the graph of the quadratic function written in standard form as \(y=−3(x+2)^2+4\). Since \(x–h=x+2\) in this example, \(h=–2\). In this form, \(a=−3\), \(h=−2\), and \(k=4\). Because \(a<0\), the parabola opens downward. The vertex is at \((−2, 4)\).

The standard form is useful for determining how the graph is transformed from the graph of \(y=x^2\). Figure \(\PageIndex{6}\) is the graph of this basic function.

If \(k>0\), the graph shifts upward, whereas if \(k<0\), the graph shifts downward. In Figure \(\PageIndex{5}\), \(k>0\), so the graph is shifted 4 units upward. If \(h>0\), the graph shifts toward the right and if \(h<0\), the graph shifts to the left. In Figure \(\PageIndex{5}\), \(h<0\), so the graph is shifted 2 units to the left. The magnitude of \(a\) indicates the stretch of the graph. If \(|a|>1\), the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if \(|a|<1\), the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure \(\PageIndex{5}\), \(|a|>1\), so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

\[\begin{align*} a(x−h)^2+k &= ax^2+bx+c \\[4pt] ax^2−2ahx+(ah^2+k)&=ax^2+bx+c \end{align*} \]

For the linear terms to be equal, the coefficients must be equal.

\[–2ah=b \text{, so } h=−\dfrac{b}{2a}. \nonumber\]

This is the axis of symmetry we defined earlier. Setting the constant terms equal:

\[\begin{align*} ah^2+k&=c \\ k&=c−ah^2 \\ &=c−a−\Big(\dfrac{b}{2a}\Big)^2 \\ &=c−\dfrac{b^2}{4a} \end{align*}\]

In practice, though, it is usually easier to remember that \(k\) is the output value of the function when the input is \(h\), so \(f(h)=k\).

Definitions: Forms of Quadratic Functions

A quadratic function is a function of degree two. The graph of a quadratic function is a parabola.

• The general form of a quadratic function is \(f(x)=ax^2+bx+c\) where \(a\), \(b\), and \(c\) are real numbers and \(a{\neq}0\).
• The standard form of a quadratic function is \(f(x)=a(x−h)^2+k\).
• The vertex \((h,k)\) is located at \[h=–\dfrac{b}{2a},\;k=f(h)=f(\dfrac{−b}{2a}).\]

HOWTO: Write a quadratic function in a general form

Given a graph of a quadratic function, write the equation of the function in general form.

• Identify the horizontal shift of the parabola; this value is \(h\). Identify the vertical shift of the parabola; this value is \(k\).
• Substitute the values of the horizontal and vertical shift for \(h\) and \(k\). in the function \(f(x)=a(x–h)^2+k\).
• Substitute the values of any point, other than the vertex, on the graph of the parabola for \(x\) and \(f(x)\).
• Solve for the stretch factor, \(|a|\).
• If the parabola opens up, \(a>0\). If the parabola opens down, \(a<0\) since this means the graph was reflected about the x-axis.
• Expand and simplify to write in general form.

Example \(\PageIndex{2}\): Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function \(g\) in Figure \(\PageIndex{7}\) as a transformation of \(f(x)=x^2\), and then expand the formula, and simplify terms to write the equation in general form.

We can see the graph of \(g\) is the graph of \(f(x)=x^2\) shifted to the left 2 and down 3, giving a formula in the form \(g(x)=a(x+2)^2–3\).

Substituting the coordinates of a point on the curve, such as \((0,−1)\), we can solve for the stretch factor.

\[\begin{align} −1&=a(0+2)^2−3 \\ 2&=4a \\ a&=\dfrac{1}{2} \end{align}\]

In standard form, the algebraic model for this graph is \(g(x)=\dfrac{1}{2}(x+2)^2–3\).

To write this in general polynomial form, we can expand the formula and simplify terms.

\[\begin{align} g(x)&=\dfrac{1}{2}(x+2)^2−3 \\ &=\dfrac{1}{2}(x+2)(x+2)−3 \\ &=\dfrac{1}{2}(x^2+4x+4)−3 \\ &=\dfrac{1}{2}x^2+2x+2−3 \\ &=\dfrac{1}{2}x^2+2x−1 \end{align}\]

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

We can check our work using the table feature on a graphing utility. First enter \(\mathrm{Y1=\dfrac{1}{2}(x+2)^2−3}\). Next, select \(\mathrm{TBLSET}\), then use \(\mathrm{TblStart=–6}\) and \(\mathrm{ΔTbl = 2}\), and select \(\mathrm{TABLE}\). See Table \(\PageIndex{1}\)

The ordered pairs in the table correspond to points on the graph.

Exercise \(\PageIndex{2}\)

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure \(\PageIndex{8}\). Find an equation for the path of the ball. Does the shooter make the basket?

Figure \(\PageIndex{8}\): Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes. (credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at \((−4, 7)\), so \(h(x)=–\frac{7}{16}(x+4)^2+7\). To make the shot, \(h(−7.5)\) would need to be about 4 but \(h(–7.5){\approx}1.64\); he doesn’t make it.

• Identify \(a\), \(b\), and \(c\).
• Find \(h\), the x-coordinate of the vertex, by substituting \(a\) and \(b\) into \(h=–\frac{b}{2a}\).
• Find \(k\), the y-coordinate of the vertex, by evaluating \(k=f(h)=f\Big(−\frac{b}{2a}\Big)\).

Example \(\PageIndex{3}\): Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function \(f(x)=2x^2–6x+7\). Rewrite the quadratic in standard form (vertex form).

The horizontal coordinate of the vertex will be at

\[\begin{align} h&=–\dfrac{b}{2a} \\ &=-\dfrac{-6}{2(2)} \\ &=\dfrac{6}{4} \\ &=\dfrac{3}{2}\end{align}\]

The vertical coordinate of the vertex will be at

\[\begin{align} k&=f(h) \\ &=f\Big(\dfrac{3}{2}\Big) \\ &=2\Big(\dfrac{3}{2}\Big)^2−6\Big(\dfrac{3}{2}\Big)+7 \\ &=\dfrac{5}{2} \end{align}\]

Rewriting into standard form, the stretch factor will be the same as the \(a\) in the original quadratic.

\[f(x)=ax^2+bx+c \\ f(x)=2x^2−6x+7\]

Using the vertex to determine the shifts,

\[f(x)=2\Big(x–\dfrac{3}{2}\Big)^2+\dfrac{5}{2}\]

One reason we may want to identify the vertex of the parabola is that this point will inform us what the maximum or minimum value of the function is, \((k)\),and where it occurs, \((h)\).

Exercise \(\PageIndex{3}\)

Given the equation \(g(x)=13+x^2−6x\), write the equation in general form and then in standard form.

\(g(x)=x^2−6x+13\) in general form; \(g(x)=(x−3)^2+4\) in standard form.

Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola . We can see the maximum and minimum values in Figure \(\PageIndex{9}\).

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Example \(\PageIndex{5}\): Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

• Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length \(L\).
• What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as Figure \(\PageIndex{10}\) to record the given information. It is also helpful to introduce a temporary variable, \(W\), to represent the width of the garden and the length of the fence section parallel to the backyard fence.

a. We know we have only 80 feet of fence available, and \(L+W+L=80\), or more simply, \(2L+W=80\). This allows us to represent the width, \(W\), in terms of \(L\).

\[W=80−2L\]

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

\[\begin{align} A&=LW=L(80−2L) \\ A(L)&=80L−2L^2 \end{align}\]

This formula represents the area of the fence in terms of the variable length \(L\). The function, written in general form, is

\[A(L)=−2L^2+80L\].

The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since \(a\) is the coefficient of the squared term, \(a=−2\), \(b=80\), and \(c=0\).

To find the vertex:

\[\begin{align} h& =−\dfrac{80}{2(−2)} &k&=A(20) \\ &=20 & \text{and} \;\;\;\; &=80(20)−2(20)^2 \\ &&&=800 \end{align}\]

The maximum value of the function is an area of 800 square feet, which occurs when \(L=20\) feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure \(\PageIndex{11}\).

• Write a quadratic equation for revenue.
• Find the vertex of the quadratic equation.
• Determine the y-value of the vertex.

Example \(\PageIndex{6}\): Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, \(p\) for price per subscription and \(Q\) for quantity, giving us the equation \(\text{Revenue}=pQ\).

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently \(p=30\) and \(Q=84,000\). We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, \(p=32\) and \(Q=79,000\). From this we can find a linear equation relating the two quantities. The slope will be

\[\begin{align} m&=\dfrac{79,000−84,000}{32−30} \\ &=−\dfrac{5,000}{2} \\ &=−2,500 \end{align}\]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

\[\begin{align} Q&=−2500p+b &\text{Substitute in the point $Q=84,000$ and $p=30$} \\ 84,000&=−2500(30)+b &\text{Solve for $b$} \\ b&=159,000 \end{align}\]

This gives us the linear equation \(Q=−2,500p+159,000\) relating cost and subscribers. We now return to our revenue equation.

\[\begin{align} \text{Revenue}&=pQ \\ \text{Revenue}&=p(−2,500p+159,000) \\ \text{Revenue}&=−2,500p^2+159,000p \end{align}\]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

\[\begin{align} h&=−\dfrac{159,000}{2(−2,500)} \\ &=31.8 \end{align}\]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

\[\begin{align} \text{maximum revenue}&=−2,500(31.8)^2+159,000(31.8) \\ &=2,528,100 \end{align}\]

This could also be solved by graphing the quadratic as in Figure \(\PageIndex{12}\). We can see the maximum revenue on a graph of the quadratic function.

Example \(\PageIndex{10}\): Applying the Vertex and x-Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation \(H(t)=−16t^2+80t+40\).

When does the ball reach the maximum height? What is the maximum height of the ball? When does the ball hit the ground?

The ball reaches the maximum height at the vertex of the parabola. \[\begin{align} h &= −\dfrac{80}{2(−16)} \\ &=\dfrac{80}{32} \\ &=\dfrac{5}{2} \\ & =2.5 \end{align}\]

The ball reaches a maximum height after 2.5 seconds.

To find the maximum height, find the y-coordinate of the vertex of the parabola. \[\begin{align} k &=H(−\dfrac{b}{2a}) \\ &=H(2.5) \\ &=−16(2.5)^2+80(2.5)+40 \\ &=140 \end{align}\]

The ball reaches a maximum height of 140 feet.

To find when the ball hits the ground, we need to determine when the height is zero, \(H(t)=0\).

We use the quadratic formula.

\[\begin{align} t & =\dfrac{−80±\sqrt{80^2−4(−16)(40)}}{2(−16)} \\ & = \dfrac{−80±\sqrt{8960}}{−32} \end{align} \]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

\[t=\dfrac{−80-\sqrt{8960}}{−32} ≈5.458 \text{ or }t=\dfrac{−80+\sqrt{8960}}{−32} ≈−0.458 \]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure \(\PageIndex{16}\).

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