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3.3: Quadratic Functions

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    114003
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    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.
    Satellite dishes.
    Figure 1 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr)

    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.

    Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are.
    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.

    Example 1

    Identifying the Characteristics of a Parabola

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

    Graph of a parabola with a vertex at (3, 1) and a y-intercept at (0, 7).
    Figure 3
    Answer

    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

    f(x)=a x 2 +bx+c f(x)=a x 2 +bx+c

    where a,b, a,b, and c c are real numbers and a0. a0. 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 2a . x= b 2a . If we use the quadratic formula, x= b± b 2 4ac 2a , x= b± b 2 4ac 2a , to solve a x 2 +bx+c=0 a x 2 +bx+c=0 for the x- x- intercepts, or zeros, we find the value of x x halfway between them is always x= b 2a , x= b 2a , the equation for the axis of symmetry.

    Figure 4 represents the graph of the quadratic function written in general form as y= x 2 +4x+3. y= x 2 +4x+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).

    Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are for the function y=x^2+4x+3.
    Figure 4

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

    f(x)=a (xh) 2 +k f(x)=a (xh) 2 +k

    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 xh=x+2 xh=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 ).

    Graph of a parabola showing where the x and y intercepts, vertex, and axis of symmetry are for the function y=-3(x+2)^2+4.
    Figure 5

    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.

    Graph of y=x^2.
    Figure 6

    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.

    a (xh) 2 +k=a x 2 +bx+c a x 2 2ahx+(a h 2 +k)=a x 2 +bx+c a (xh) 2 +k=a x 2 +bx+c a x 2 2ahx+(a h 2 +k)=a x 2 +bx+c

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

    –2ah=b, so h= b 2a . –2ah=b, so h= b 2a .

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

    a h 2 +k=c k=ca h 2 =ca( b 2a ) 2 =c b 2 4a a h 2 +k=c k=ca h 2 =ca( b 2a ) 2 =c b 2 4a

    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 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 +bx+c f(x)=a x 2 +bx+c where a,b, a,b, and c c are real numbers and a0. a0.

    The standard form of a quadratic function is f(x)=a (xh) 2 +k. f(x)=a (xh) 2 +k.

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

    h= b 2a ,k=f(h)=f( b 2a ). h= b 2a ,k=f(h)=f( b 2a ).

    How To

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

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

    Example 2

    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.

    Graph of a parabola with its vertex at (-2, -3).
    Figure 7
    Answer

    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. g(x)=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.

    1=a (0+2) 2 3 2=4a a= 1 2 1=a (0+2) 2 3 2=4a a= 1 2

    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.

    g(x)= 1 2 (x+2) 2 3 = 1 2 (x+2)(x+2)3 = 1 2 ( x 2 +4x+4)3 = 1 2 x 2 +2x+23 = 1 2 x 2 +2x1 g(x)= 1 2 (x+2) 2 3 = 1 2 (x+2)(x+2)3 = 1 2 ( x 2 +4x+4)3 = 1 2 x 2 +2x+23 = 1 2 x 2 +2x1

    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.

    Analysis

    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 ΔTbl = 2, ΔTbl = 2, and select TABLE. TABLE. See Table 1.

    x x –6 –4 –2 0 2
    y y 5 –1 –3 –1 5
    Table 1

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

    Try It #1

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

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

    How To

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

    1. Identify a,b,andc. a,b,andc.
    2. Find h, h, the x-coordinate of the vertex, by substituting a a and b b into h= b 2a . h= b 2a .
    3. Find k, k, the y-coordinate of the vertex, by evaluating k=f( h )=f( b 2a ). k=f( h )=f( b 2a ).

    Example 3

    Finding the Vertex of a Quadratic Function

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

    Answer

    The horizontal coordinate of the vertex will be at

    h= b 2a = 6 2(2) = 6 4 = 3 2 h= b 2a = 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 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.

    f(x)=a x 2 +bx+c f(x)=2 x 2 6x+7 f(x)=a x 2 +bx+c f(x)=2 x 2 6x+7

    Using the vertex to determine the shifts,

    f( x )=2 ( x 3 2 ) 2 + 5 2 f( x )=2 ( x 3 2 ) 2 + 5 2

    Analysis

    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 ).

    Try It #2

    Given the equation g(x)=13+ x 2 6x, g(x)=13+ x 2 6x, 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.

    The range of a quadratic function written in general form f(x)=a x 2 +bx+c f(x)=a x 2 +bx+c with a positive a a value is f(x)f( b 2a ), f(x)f( b 2a ), or [ f( b 2a ), ). [ f( b 2a ), ).

    The range of a quadratic function written in general form with a negative a a value is f(x)f( b 2a ), f(x)f( b 2a ), or ( ,f( b 2a ) ]. ( ,f( b 2a ) ].

    The range of a quadratic function written in standard form f(x)=a (xh) 2 +k f(x)=a (xh) 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.

    How To

    Given a quadratic function, find the domain and range.

    1. Identify the domain of any quadratic function as all real numbers.
    2. 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.
    3. Determine the maximum or minimum value of the parabola, k. k.
    4. 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 ].

    Example 4

    Finding the Domain and Range of a Quadratic Function

    Find the domain and range of f(x)=5 x 2 +9x1. f(x)=5 x 2 +9x1.

    Answer

    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.

    h= b 2a = 9 2(5) = 9 10 h= b 2a = 9 2(5) = 9 10

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

    f( 9 10 )=5 ( 9 10 ) 2 +9( 9 10 )1 = 61 20 f( 9 10 )=5 ( 9 10 ) 2 +9( 9 10 )1 = 61 20

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

    Try It #3

    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.

    Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).
    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.

    Example 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.

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

    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.

    Diagram of the garden and the backyard.
    Figure 10
    1. W=802L W=802L

      A=LW=L(802L) A(L)=80L2 L 2 A=LW=L(802L) A(L)=80L2 L 2

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

      A(L)=2 L 2 +80L. A(L)=2 L 2 +80L.

    2. 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:

    h= 80 2(2) k=A(20) =20 and =80(20)2 (20) 2 =800 h= 80 2(2) k=A(20) =20 and =80(20)2 (20) 2 =800

    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.

    Analysis

    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.

    Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).
    Figure 11

    How To

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

    1. Write a quadratic equation for revenue.
    2. Find the vertex of the quadratic equation.
    3. Determine the y-value of the vertex.

    Example 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?

    Answer

    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=pQ. 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 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

    m= 79,00084,000 3230 = 5,000 2 =2,500 m= 79,00084,000 3230 = 5,000 2 =2,500

    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.

    Q=−2500p+b Substitute in the point Q=84,000 and p=30 84,000=−2500(30)+b Solve for b b=159,000 Q=−2500p+b Substitute in the point Q=84,000 and p=30 84,000=−2500(30)+b Solve for b b=159,000

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

    Revenue=pQ Revenue=p(−2,500p+159,000) Revenue=−2,500 p 2 +159,000p Revenue=pQ Revenue=p(−2,500p+159,000) Revenue=−2,500 p 2 +159,000p

    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.

    h= 159,000 2(2,500) =31.8 h= 159,000 2(2,500) =31.8

    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.

    maximum revenue=−2,500 (31.8) 2 +159,000(31.8) =2,528,100 maximum revenue=−2,500 (31.8) 2 +159,000(31.8) =2,528,100

    Analysis

    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.

    Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).
    Figure 12

    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.

    Three graphs where the first graph shows a parabola with no x-intercept, the second is a parabola with one –intercept, and the third parabola is of two x-intercepts.
    Figure 13 Number of x-intercepts of a parabola

    How To

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

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

    Example 7

    Finding the y- and x-Intercepts of a Parabola

    Find the y- and x-intercepts of the quadratic f(x)=3 x 2 +5x2. f(x)=3 x 2 +5x2.

    Answer

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

    f(0)=3 (0) 2 +5(0)2 =2 f(0)=3 (0) 2 +5(0)2 =2

    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.

    0=3 x 2 +5x2 0=3 x 2 +5x2

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

    0=(3x1)(x+2) 0=(3x1)(x+2)

    0=3x1 0=x+2 x= 1 3 or x=2 0=3x1 0=x+2 x= 1 3 or x=2

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

    Analysis

    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

    Graph of a parabola which has the following intercepts (-2, 0), (1/3, 0), and (0, -2).
    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.

    How To

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

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

    Example 8

    Finding the x- x- Intercepts of a Parabola

    Find the x- x- intercepts of the quadratic function f(x)=2 x 2 +4x4. f(x)=2 x 2 +4x4.

    Answer

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

    0=2 x 2 +4x4 0=2 x 2 +4x4

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

    f( x )=a ( xh ) 2 +k f( x )=a ( xh ) 2 +k

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

    h= b 2a k=f(1) = 4 2(2) =2 (1) 2 +4(1)4 =−1 =−6 h= b 2a k=f(1) = 4 2(2) =2 (1) 2 +4(1)4 =−1 =−6

    So now we can rewrite in standard form.

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

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

    0=2 (x+1) 2 6 6=2 (x+1) 2 3= (x+1) 2 x+1=± 3 x=1± 3 0=2 (x+1) 2 6 6=2 (x+1) 2 3= (x+1) 2 x+1=± 3 x=1± 3

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

    Analysis

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

    Graph of a parabola which has the following x-intercepts (-2.732, 0) and (0.732, 0).
    Figure 15

    Try It #4

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

    Example 9

    Solving a Quadratic Equation with the Quadratic Formula

    Solve x 2 +x+2=0. x 2 +x+2=0.

    Answer

    Let’s begin by writing the quadratic formula: x= b± b 2 4ac 2a . x= b± b 2 4ac 2a .

    When applying the quadratic formula, we identify the coefficients a,b and c. a,b and c. For the equation x 2 +x+2=0, x 2 +x+2=0, we have a=1,b=1,andc=2. a=1,b=1,andc=2. Substituting these values into the formula we have:

    x= b± b 2 4ac 2a = 1± 1 2 41(2) 21 = 1± 18 2 = 1± 7 2 = 1±i 7 2 x= b± b 2 4ac 2a = 1± 1 2 41(2) 21 = 1± 18 2 = 1± 7 2 = 1±i 7 2

    The solutions to the equation are 1+i 7 2 1+i 7 2 and 1i 7 2 1i 7 2 or 1 2 + i 7 2 1 2 + i 7 2 and 1 2 i 7 2 . 1 2 i 7 2 .

    Example 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)=16 t 2 +80t+40. H(t)=16 t 2 +80t+40.

    1. When does the ball reach the maximum height?
    2. What is the maximum height of the ball?
    3. When does the ball hit the ground?
    Answer

    1. The ball reaches the maximum height at the vertex of the parabola.

      h= 80 2(16) = 80 32 = 5 2 =2.5 h= 80 2(16) = 80 32 = 5 2 =2.5

      The ball reaches a maximum height after 2.5 seconds.

    2. To find the maximum height, find the y- y- coordinate of the vertex of the parabola.

      k=H( b 2a ) =H( 2.5 ) =−16 ( 2.5 ) 2 +80( 2.5 )+40 =140 k=H( b 2a ) =H( 2.5 ) =−16 ( 2.5 ) 2 +80( 2.5 )+40 =140

      The ball reaches a maximum height of 140 feet.

    3. We use the quadratic formula.

      t= 80± 80 2 4(16)(40) 2(16) = 80± 8960 32 t= 80± 80 2 4(16)(40) 2(16) = 80± 8960 32

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

      t= 80 8960 32 5.458 or t= 80+ 8960 32 0.458 t= 80 8960 32 5.458 or t= 80+ 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 16

      Graph of a negative parabola where x goes from -1 to 6.
      Figure 16

    Try It #5

    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 +96t+112. H(t)=16 t 2 +96t+112.

    1. When does the rock reach the maximum height?
    2. What is the maximum height of the rock?
    3. When does the rock hit the ocean?

    Media

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

    3.2 Section Exercises

    Verbal

    1.

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

    2.

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

    3.

    Explain why the condition of a0 a0 is imposed in the definition of the quadratic function.

    4.

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

    5.

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

    Algebraic

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

    6.

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

    7.

    g( x )= x 2 +2x3 g( x )= x 2 +2x3

    8.

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

    9.

    f(x)= x 2 +5x2 f(x)= x 2 +5x2

    10.

    h( x )=2 x 2 +8x10 h( x )=2 x 2 +8x10

    11.

    k( x )=3 x 2 6x9 k( x )=3 x 2 6x9

    12.

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

    13.

    f(x)=3 x 2 5x1 f(x)=3 x 2 5x1

    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.

    14.

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

    15.

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

    16.

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

    17.

    f(x)=4 x 2 +x1 f(x)=4 x 2 +x1

    18.

    h( t )=4 t 2 +6t1 h( t )=4 t 2 +6t1

    19.

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

    20.

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

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

    21.

    f(x)= (x3) 2 +2 f(x)= (x3) 2 +2

    22.

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

    23.

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

    24.

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

    25.

    k( x )=3 x 2 6x9 k( x )=3 x 2 6x9

    For the following exercises, solve the equations over the complex numbers.

    26.

    x 2 =25 x 2 =25

    27.

    x 2 =8 x 2 =8

    28.

    x 2 +36=0 x 2 +36=0

    29.

    x 2 +27=0 x 2 +27=0

    30.

    x 2 +2x+5=0 x 2 +2x+5=0

    31.

    x 2 4x+5=0 x 2 4x+5=0

    32.

    x 2 +8x+25=0 x 2 +8x+25=0

    33.

    x 2 4x+13=0 x 2 4x+13=0

    34.

    x 2 +6x+25=0 x 2 +6x+25=0

    35.

    x 2 10x+26=0 x 2 10x+26=0

    36.

    x 2 6x+10=0 x 2 6x+10=0

    37.

    x(x4)=20 x(x4)=20

    38.

    x(x2)=10 x(x2)=10

    39.

    2 x 2 +2x+5=0 2 x 2 +2x+5=0

    40.

    5 x 2 8x+5=0 5 x 2 8x+5=0

    41.

    5 x 2 +6x+2=0 5 x 2 +6x+2=0

    42.

    2 x 2 6x+5=0 2 x 2 6x+5=0

    43.

    x 2 +x+2=0 x 2 +x+2=0

    44.

    x 2 2x+4=0 x 2 2x+4=0

    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.

    45.

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

    46.

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

    47.

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

    48.

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

    49.

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

    50.

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

    51.

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

    52.

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

    Graphical

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

    53.

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

    54.

    f(x)= x 2 6x1 f(x)= x 2 6x1

    55.

    f(x)= x 2 5x6 f(x)= x 2 5x6

    56.

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

    57.

    f(x)=2 x 2 +5x8 f(x)=2 x 2 +5x8

    58.

    f(x)=4 x 2 12x3 f(x)=4 x 2 12x3

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

    59.
    Graph of a positive parabola with a vertex at (2, -3) and y-intercept at (0, 1).
    60.
    Graph of a positive parabola with a vertex at (-1, 2) and y-intercept at (0, 3)
    61.
    Graph of a negative parabola with a vertex at (2, 7).
    62.
    Graph of a negative parabola with a vertex at (-1, 2).
    63.
    Graph of a positive parabola with a vertex at (3, -1) and y-intercept at (0, 3.5).
    64.
    Graph of a negative parabola with a vertex at (-2, 3).

    Numeric

    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.

    65.
    x x –2 –1 0 1 2
    y y 5 2 1 2 5
    66.
    x x –2 –1 0 1 2
    y y 1 0 1 4 9
    67.
    x x –2 –1 0 1 2
    y y –2 1 2 1 –2
    68.
    x x –2 –1 0 1 2
    y y –8 –3 0 1 0
    69.
    x x –2 –1 0 1 2
    y y 8 2 0 2 8

    Technology

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

    70.

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

    What appears to be the effect of changing the coefficient?

    71.

    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?

    72.

    Graph on the same set of axes f(x)= x 2 f(x)= x 2 ,f(x)= (x2) 2 f(x)= (x2) 2 ,f (x3) 2 f (x3) 2 , and f(x)= (x+4) 2 . and f(x)= (x+4) 2 .

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

    73.

    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.

    74.

    A suspension bridge can be modeled by the quadratic function h(x)=.0001 x 2 h(x)=.0001 x 2 with 2000x2000 2000x2000 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.

    Extensions

    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.

    75.

    Vertex (1,−2), (1,−2), opens up.

    76.

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

    77.

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

    78.

    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.

    79.

    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.

    80.

    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.

    81.

    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.

    82.

    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.

    83.

    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.

    84.

    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

    85.

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

    86.

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

    87.

    Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing.

    88.

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

    89.

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

    90.

    Suppose that the price per unit in dollars of a cell phone production is modeled by p=$450.0125x, p=$450.0125x, where x x is in thousands of phones produced, and the revenue represented by thousands of dollars is R=xp. R=xp. Find the production level that will maximize revenue.

    91.

    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 +229t+234. h( t )=4.9 t 2 +229t+234. Find the maximum height the rocket attains.

    92.

    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 +24t+8. h( t )=4.9 t 2 +24t+8. How long does it take to reach maximum height?

    93.

    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?

    94.

    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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