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.

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.

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 x𝑥 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.

Figure 3

Solution

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)=ax2+bx+c𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐

where a,b,𝑎,𝑏, and c𝑐 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=−b2a.𝑥=−𝑏2𝑎. If we use the quadratic formula, x=−b±b2−4ac√2a,𝑥=−𝑏±𝑏2−4𝑎𝑐2𝑎, to solve ax2+bx+c=0𝑎𝑥2+𝑏𝑥+𝑐=0 for the x-𝑥- intercepts, or zeros, we find the value of x𝑥 halfway between them is always x=−b2a,𝑥=−𝑏2𝑎, the equation for the axis of symmetry.

Figure 4 represents the graph of the quadratic function written in general form as y=x2+4x+3.𝑦=𝑥2+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=−42(1)=−2.𝑥=−42(1)=−2. This also makes sense because we can see from the graph that the vertical line x=−2𝑥=−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).

Figure 4

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

f(x)=a(x−h)2+k𝑓(𝑥)=𝑎(𝑥−ℎ)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 x–h=x+2𝑥–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).

Figure 5

The standard form is useful for determining how the graph is transformed from the graph of y=x2.𝑦=𝑥2. Figure 6 is the graph of this basic function.

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 a𝑎 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(x−h)2+kax2−2ahx+(ah2+k)==ax2+bx+cax2+bx+c𝑎(𝑥−ℎ)2+𝑘=𝑎𝑥2+𝑏𝑥+𝑐𝑎𝑥2−2𝑎ℎ𝑥+(𝑎ℎ2+𝑘)=𝑎𝑥2+𝑏𝑥+𝑐

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

–2ah=b,so h=−b2a–2𝑎ℎ=𝑏,so ℎ=−𝑏2𝑎

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

ah2+kk====cc−ah2c−a−(b2a)2c−b24a𝑎ℎ2+𝑘=𝑐𝑘=𝑐−𝑎ℎ2=𝑐−𝑎−(𝑏2𝑎)2=𝑐−𝑏24𝑎

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)=ax2+bx+c𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐 where a,b,𝑎,𝑏, and c𝑐 are real numbers and a≠0.𝑎≠0.

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

The vertex (h,k)(ℎ,𝑘) is located at

h=–b2a,k=f(h)=f(−b2a)ℎ=–𝑏2𝑎,𝑘=𝑓(ℎ)=𝑓(−𝑏2𝑎)

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 hℎ and k.𝑘. in the function f(x)=a(x–h)2+k.𝑓(𝑥)=𝑎(𝑥–ℎ)2+𝑘.
3. Substitute the values of any point, other than the vertex, on the graph of the parabola for x𝑥 and f(x).𝑓(𝑥).
4. Solve for the stretch factor, |a|.|𝑎|.
5. 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 g𝑔 in Figure 7 as a transformation of f(x)=x2,𝑓(𝑥)=𝑥2, and then expand the formula, and simplify terms to write the equation in general form.

Figure 7

Solution

We can see the graph of g is the graph of f(x)=x2𝑓(𝑥)=𝑥2 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.

−12a===a(0+2)2−34a12−1=𝑎(0+2)2−32=4𝑎𝑎=12

In standard form, the algebraic model for this graph is (g)x=12(x+2)2–3.(𝑔)𝑥=12(𝑥+2)2–3.

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

g(x)=====12(x+2)2−312(x+2)(x+2)−312(x2+4x+4)−312×2+2x+2−312×2+2x−1𝑔(𝑥)=12(𝑥+2)2−3=12(𝑥+2)(𝑥+2)−3=12(𝑥2+4𝑥+4)−3=12𝑥2+2𝑥+2−3=12𝑥2+2𝑥−1

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=12(x+2)2−3.Y1=12(𝑥+2)2−3. Next, select TBLSET,TBLSET, then use TblStart=–6TblStart=–6 and ΔTbl = 2,𝛥Tbl = 2, and select TABLE.TABLE. See Table 1.

x𝑥	–6	–4	–2	0	2
y𝑦	5	–1	–3	–1	5

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. 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?

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, and c.𝑎, 𝑏, and 𝑐.
2. Find h,ℎ, the x-coordinate of the vertex, by substituting a𝑎 and b𝑏 into h=–b2a.ℎ=–𝑏2𝑎.
3. Find k,𝑘, the y-coordinate of the vertex, by evaluating k=f(h)=f(−b2a).𝑘=𝑓(ℎ)=𝑓(−𝑏2𝑎).

EXAMPLE 3

Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function f(x)=2×2–6x+7.𝑓(𝑥)=2𝑥2–6𝑥+7. Rewrite the quadratic in standard form (vertex form).

Solution

The horizontal coordinate of the vertex will be atThe vertical coordinate of the vertex will be athk========−b2a−−62(2)6432f(h)f(32)2(32)2−6(32)+752The horizontal coordinate of the vertex will be atℎ=−𝑏2𝑎=−−62(2)=64=32The vertical coordinate of the vertex will be at𝑘=𝑓(ℎ)=𝑓(32)=2(32)2−6(32)+7=52

Rewriting into standard form, the stretch factor will be the same as the a𝑎 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 ” a𝑎 ” from the general form.

f(x)f(x)==ax2+bx+c2x2−6x+7𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐𝑓(𝑥)=2𝑥2−6𝑥+7

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

f(x)=2(x–32)2+52𝑓(𝑥)=2(𝑥–32)2+52

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+x2−6x,𝑔(𝑥)=13+𝑥2−6𝑥, 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 unless the context of the function presents some restrictions.

The range of a quadratic function written in general form f(x)=ax2+bx+c𝑓(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐 with a positive a𝑎 value is f(x)≥f(−b2a),𝑓(𝑥)≥𝑓(−𝑏2𝑎), or [f(−b2a),∞);[𝑓(−𝑏2𝑎),∞); the range of a quadratic function written in general form with a negative a𝑎 value is f(x)≤f(−b2a),𝑓(𝑥)≤𝑓(−𝑏2𝑎), or (−∞,f(−b2a)].(−∞,𝑓(−𝑏2𝑎)].

The range of a quadratic function written in standard form f(x)=a(x−h)2+k𝑓(𝑥)=𝑎(𝑥−ℎ)2+𝑘 with a positive a𝑎 value is f(x)≥k;𝑓(𝑥)≥𝑘; the range of a quadratic function written in standard form with a negative a𝑎 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 a𝑎 is positive or negative. If a𝑎 is positive, the parabola has a minimum. If a𝑎 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)=−5×2+9x−1.𝑓(𝑥)=−5𝑥2+9𝑥−1.

Solution

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

Because a𝑎 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===−b2a−92(−5)910ℎ=−𝑏2𝑎=−92(−5)=910

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

f(910)==−5(910)2+9(910)−16120𝑓(910)=-5(910)2+9(910)−1=6120

The range is f(x)≤6120,𝑓(𝑥)≤6120, or (−∞,6120].(−∞,6120].

TRY IT #3

Find the domain and range of f(x)=2(x−47)2+811.𝑓(𝑥)=2(𝑥−47)2+811.

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.

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?

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.

Figure 11

HOW TO

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

1. Write a quadratic equation for a revenue function.
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?

Solution

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, p𝑝 for price per subscription and Q𝑄 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=30𝑝=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=32𝑝=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===79,000−84,00032−30−5,0002−2,500𝑚=79,000−84,00032−30=−5,0002=−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.

Q84,000b===−2500p+b−2500(30)+b159,000Substitute in the pointQ=84,000and p=30Solve forb𝑄=−2500𝑝+𝑏Substitute in the point𝑄=84,000and 𝑝=3084,000=−2500(30)+𝑏Solve for𝑏𝑏=159,000

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

RevenueRevenueRevenue===pQp(−2,500p+159,000)−2,500p2+159,000pRevenue=𝑝𝑄Revenue=𝑝(−2,500𝑝+159,000)Revenue=−2,500𝑝2+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==−159,0002(−2,500)31.8ℎ=−159,0002(−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)2,528,100maximum 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.

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.

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-intercept.
2. Solve the quadratic equation f(x)=0𝑓(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)=3×2+5x−2.𝑓(𝑥)=3𝑥2+5𝑥−2.

Solution

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

f(0)==3(0)2+5(0)−2−2𝑓(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=3×2+5x−20=3𝑥2+5𝑥−2

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

0=(3x−1)(x+2)0=(3𝑥−1)(𝑥+2)

So the x-intercepts are at (13,0)(13,0) 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 (13,0)(13,0) and (−2,0).(−2,0). See Figure 14

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 a𝑎 and b𝑏 into h=−b2a.ℎ=−𝑏2𝑎.
2. Substitute x=h𝑥=ℎ into the general form of the quadratic function to find k.𝑘.
3. Rewrite the quadratic in standard form using hℎ 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-Intercepts of a Parabola

Find the x-𝑥- intercepts of the quadratic function f(x)=2×2+4x−4.𝑓(𝑥)=2𝑥2+4𝑥−4.

Solution

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

0=2×2+4x−40=2𝑥2+4𝑥−4

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(x−h)2+k𝑓(𝑥)=𝑎(𝑥−ℎ)2+𝑘

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

h===−b2a−42(2)−1k===f(−1)2(−1)2+4(−1)−4−6ℎ=−𝑏2𝑎𝑘=𝑓(−1)=−42(2)=2(−1)2+4(−1)−4=−1=−6

So now we can rewrite in standard form.

f(x)=2(x+1)2−6𝑓(𝑥)=2(𝑥+1)2−6

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

0=2(x+1)2−66=2(x+1)23=(x+1)2x+1=±3–√x=−1±3–√0=2(𝑥+1)2−66=2(𝑥+1)23=(𝑥+1)2𝑥+1=±3𝑥=−1±3

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.

Figure 15

Analysis

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

x=====−b±b2−4ac√2a−4±42−4(2)(−4)√2(2)−4±48√4−4±3(16)√4−1±3–√𝑥=−𝑏±𝑏2−4𝑎𝑐2𝑎=−4±42−4(2)(−4)2(2)=−4±484=−4±3(16)4=−1±3

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

TRY IT #4

In a Try It, we found the standard and general form for the function g(x)=13+x2−6x.𝑔(𝑥)=13+𝑥2−6𝑥. Now find the y– and x-intercepts (if any).

EXAMPLE 9

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)=−16t2+80t+40.𝐻(𝑡)=−16𝑡2+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?

Solution

1. ⓐThe ball reaches the maximum height at the vertex of the parabola.h====−802(−16)8032522.5ℎ=−802(−16)=8032=52=2.5The 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(−b2a)H(2.5)−16(2.5)2+80(2.5)+40140𝑘=𝐻(−𝑏2𝑎)=𝐻(2.5)=−16(2.5)2+80(2.5)+40=140The ball reaches a maximum height of 140 feet.
3. ⓒTo find when the ball hits the ground, we need to determine when the height is zero, H(t)=0.𝐻(𝑡)=0.We use the quadratic formula.t==−80±802−4(−16)(40)√2(−16)−80±8960√−32𝑡=−80±802−4(−16)(40)2(−16)=−80±8960−32Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.t=−80−8960√−32≈5.458ort=−80+8960√−32≈−0.458𝑡=−80−8960−32≈5.458or𝑡=−80+8960−32≈−0.458The 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.Figure 16Note 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.

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)=−16t2+96t+112.𝐻(𝑡)=−16𝑡2+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?