5.3 Zeros of the Quadratic 461

Section 5.3 Zeros of the Quadratic 461

5.3 Zeros of the Quadratic We’ve seen how vertex form and intelligent use of the axis of symmetry can help to draw an accurate graph of the quadratic function defined by the equation f(x) = ax2 +bx+c. When drawing the graph of the parabola it is helpful to know where the graph of the parabola crosses the x-axis. That is the primary goal of this section, to find the zero crossings or x-intercepts of the parabola. Before we begin, you’ll need to review the techniques that will enable you to factor the quadratic expression ax2 + bx + c. Factoring ax2 + bx + c when a = 1 Our intent in this section is to provide a quick review of techniques used to factor quadratic trinomials. We begin by showing how to factor trinomials having the form ax2 + bx + c, where the leading coefficient is a = 1; that is, trinomials having the form x2 +bx+c. In the next section, we will address the technique used to factor ax2 +bx+c when a 6= 1. Let’s begin with an example.

2 I Example 1. Factor x + 16x − 36. Note that the leading coefficient, the coefficient of x2, is a 1. This is an important observation, because the technique presented here will not work when the leading coefficient does not equal 1. Note the constant term of the trinomial x2 + 16x − 36 is −36. List all integer pairs whose product equals −36.

1, −36 −1, 36 2, −18 −2, 18 3, −12 −3, 12 4, −9 −4, 9 6, −6 −6, 6

Note that we’ve framed the pair −2, 18. We’ve done this because the sum of this pair of integers equals the coeﬃcient of x in the trinomial expression x2 + 16x − 36. Use this framed pair to factor the trinomial.

x2 + 16x − 36 = (x − 2)(x + 18)

It is important that you check your result. Use the distributive property to multiply. (x − 2)(x + 18) = x(x + 18) − 2(x + 18) = x2 + 18x − 2x − 36 = x2 + 16x − 36

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Thus, our factorization is correct.

Let’s summarize the technique.

Algorithm. To factor the quadratic x2 + bx + c, proceed as follows:

1. List all the integer pairs whose product equals c. 2. Circle or frame the pair whose sum equals the coeﬃcient of x, namely b. Use this pair to factor the trinomial.

Let’s look at another example.

2 I Example 2. Factor the trinomial x − 25x − 84. List all the integer pairs whose product is −84.

1, −84 −1, 84 2, −42 −2, 42 3, −28 −3, 28 4, −21 −4, 21 6, −14 −6, 14 7, −12 −7, 12

We’ve framed the pair whose sum equals the coeﬃcient of x, namely −25. Use this pair to factor the trinomial.

x2 − 25x − 84 = (x + 3)(x − 28)

Check. (x + 3)(x − 28) = x(x − 28) + 3(x − 28) = x2 − 28x + 3x − 84 = x2 − 25x − 84

With experience, there are a number of ideas that will quicken the process.

• As you are listing the integer pairs, should you happen to note that the current pair has the appropriate sum, there is no need to list the remaining integer pairs. Simply halt the process of listing the integer pairs and use the current pair to factor the trinomial. • Some students are perfectly happy being asked “Can you think of an integer pair whose product is c and whose sum is b (where b and c refer to the coeﬃcients of x2 + bx + c)?” If you can pick the pair “out of the air” like this, all is well and good.

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Use the integer pair to factor the trinomial and don’t bother listing any integer pairs.

Now, let’s investigate how to proceed when the leading coeﬃcient is not 1. Factoring ax2 + bx + c when a 6= 1

When a 6= 1, we use a technique called the ac-test to factor the trinomial ax2 + bx + c. The process is best explained with an example.

2 I Example 3. Factor 2x + 13x − 24. Note that the leading coeﬃcient does not equal 1. Indeed, the coeﬃcient of x2 in this example is a 2. Therefore, the technique of the previous examples will not work. Thus, we turn to a similar technique called the ac-test. First, compare

2x2 + 13x − 24 and ax2 + bx + c and note that a = 2, b = 13, and c = −24. Compute the product of a and c. This is how the technique earns its name “ac-test.”

ac = (2)(−24) = −48

List all integer pairs whose product is ac = −48.

1, −48 −1, 48 2, −24 −2, 24 3, −16 −3, 16 4, −12 −4, 12 6, −8 −6, 8

We’ve framed the pair whose sum is b = 13. The next step is to rewrite the trinomial 2x2 + 13x − 24, splitting the middle term into a sum, using our framed integer pair.

2x2 + 13x − 24 = 2x2 − 3x + 16x − 24

We factor an x out of the ﬁrst two terms, then an 8 out of the last two terms. This process is called factoring by grouping.

2x2 − 3x + 16x − 24 = x(2x − 3) + 8(2x − 3)

We now factor out a common factor of 2x − 3.

x(2x − 3) + 8(2x − 3) = (x + 8)(2x − 3)

It’s helpful to see the complete process as a coherent unit.

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2x2 + 13x − 24 = 2x2 − 3x + 16x − 24 = x(2x − 3) + 8(2x − 3) = (x + 8)(2x − 3)

Check. Again, it is important to check the answer by multiplication. (x + 8)(2x − 3) = x(2x − 3) + 8(2x − 3) = 2x2 − 3x + 16x − 24 = 2x2 + 13x − 24

Because this is the original trinomial, our solution checks.2

Let’s summarize this process.

Algorithm: ac-Test. To factor the quadratic ax2 + bx + c, proceed as follows:

1. List all integer pairs whose product equals ac. 2. Circle or frame the pair whose sum equals the coeﬃcient of x, namely b. 3. Use the circled pair to express the middle term bx as a sum. 4. Factor by “grouping.”

Let’s look at another example.

2 I Example 4. Factor 3x + 34x − 24. Compare

3x2 + 34x − 24 and ax2 + bx + c

and note that a = 3, b = 34 and c = −24. List all integer pairs whose product equals ac = (3)(−24) = −72.

1, −72 −1, 72 2, −36 −2, 36 3, −24 −3, 24 4, −18 −4, 18 6, −12 −6, 12 8, −9 −8, 9

We’ve framed the pair whose sum is the same as b = 34, the coeﬃcient of x in 3x2 + 34x − 24. Again, possible shortcuts are possible. If you can “think” of a pair whose product is ac = −72 and whose sum is b = 34, then it is not necessary to list any integer pairs. Alternatively, if you come across the needed pair as you are listing

2 If you check a number of your results, it will soon become apparent why the ac-test works so well.

Version: Fall 2007 Section 5.3 Zeros of the Quadratic 465 them, then you can halt the process. There is no need to list the remaining pairs if you have the one you need. Use the framed pair to express the middle term as a sum, then factor by grouping. 3x2 + 34x − 24 = 3x2 − 2x + 36x − 24 = x(3x − 2) + 12(3x − 2) = (x + 12)(3x − 2)

We leave it to the reader to check this result.

Intercepts

The points where the graph of a function crosses the x-axis are called the x-intercepts of graph of the function. Consider the graph of the quadratic function f in Figure 1. y 10 f

x (−3, 0) (2, 0) 10

(0, −6)

Figure 1. The x- and y-intercepts are key features of any graph.

Note that the graph of the f crosses the x-axis at (−3, 0) and (2, 0). These are the x-intercepts of the parabola. Note that the y-coordinate of each x-intercept is zero. In function notation, the solutions of f(x) = 0 (note the similarity to y = 0) are the x-coordinates of the points where the graph of f crosses the x-axis. Analyzing the graph of f in Figure 1, we see that both −3 and 2 are solutions of f(x) = 0. Thus, the process for ﬁnding the x-intercepts is clear.

Finding x-intercepts. To ﬁnd the x-intercepts of the graph of any function, set y = 0 and solve for x. Alternatively, if function notation is used, set f(x) = 0 and solve for x.

Let’s look at an example.

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I Example 5. Find the x-intercepts of the graph of the quadratic function deﬁned by y = x2 + 2x − 48.

To ﬁnd the x-intercepts, ﬁrst set y = 0.

0 = x2 + 2x − 48

Next, factor the trinomial on the right. Note that the coeﬃcient of x2 is 1. We need only think of two integers whose product equals the constant term −48 and whose sum equals the coeﬃcient of x, namely 2. The numbers 8 and −6 come to mind, so the trinomial factors as follows (readers should check this result).

0 = (x + 8)(x − 6)

To complete the solution, we need to use an important property of the real numbers called the zero product property.

Zero Product Property. If a and b are any real numbers such that

ab = 0,

then either a = 0 or b = 0.

In our case, we have 0 = (x + 8)(x − 6). Therefore, it must be the case that either

x + 8 = 0 or x − 6 = 0.

These equations can be solved independently to produce

x = −8 or x = 6.

Thus, the x-intercepts of the graph of y = x2 +2x−48 are located at (−8, 0) and (6, 0).

Let’s look at another example.

I Example 6. Find the x-intercepts of the graph of the quadratic function f(x) = 2x2 − 7x − 15.

To ﬁnd the x-intercepts of the graph of the quadratic function f, we begin by setting

f(x) = 0.

Of course, f(x) = 2x2 − 7x − 15, so we can substitute to obtain

2x2 − 7x − 15 = 0.

We will now use the ac-test to factor the trinomial on the left. Note that ac = (2)(−15) = −30. List the integer pairs whose products equal −30.

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1, −30 −1, 20 2, −15 −2, 15 3, −10 −3, 10 5, −6 −5, 6

Note that the framed pair sum to the coeﬃcient of x in 2x2 − 7x − 15. Use the framed pair to express the middle term as a sum, then factor by grouping. 2x2 − 7x − 15 = 0 2x2 + 3x − 10x − 15 = 0 x(2x + 3) − 5(2x + 3) = 0 (x − 5)(2x + 3) = 0

Now we can use the zero product property. Either

x − 5 = 0 or 2x + 3 = 0.

Each of these can be solved independently to obtain

x = 5 or x = −3/2.

Thus, the x-intercepts of the graph of the quadratic function f(x) = 2x2 − 7x − 15 are located at (−3/2, 0) and (5, 0).

One more deﬁnition is in order.

Deﬁnition 7. Zeros of a Function. The solutions of f(x) = 0 are called the zeros of the function f.

Thus, in the last example, both −3/2 and 5 are zeros of the quadratic function f(x) = 2x2 −7x−15. Note the intimate relationship between the zeros of the quadratic function and the x-intercepts of the graph. Note that −3/2 is a zero and (−3/2, 0) is an x-intercept. Similarly, 5 is a zero and (5, 0) is an x-intercept. The graphing calculator can be used to ﬁnd the zeros of a function.

I Example 8. Use the graphing calculator to ﬁnd the zeros of the function f(x) = 2x2 − 7x − 15.

Enter the function f(x) = 2x2 − 7x − 15 into Y1 in the Y= menu; then adjust the window parameters as shown in Figure 2(b). Push the GRAPH button to produce the parabola shown in Figure 2(c). To ﬁnd a zero of the function, proceed as follows:

Version: Fall 2007 468 Chapter 5 Quadratic Functions

(a) (b) (c) Figure 2. Plotting the quadratic function f(x) = 2x2 − 7x − 15.

• Press 2nd TRACE to open the CALCULATE window shown in Figure 3(a). From this menu, select 2:zero. • The calculator responds by asking for a “Left bound.” Use the arrow keys to move the cursor slightly to the left of the leftmost x-intercept, as shown in Figure 3(b). Press the ENTER key. • The calculator responds by asking for a “Right bound.” Use the arrow keys to move the cursor slightly to the right of the leftmost x-intercept, as shown in Figure 3(c). Press the ENTER key. • The calculator responds by asking for a “Guess.” You may use the arrow keys to select a starting x-value any where between the left- and right-bounds you selected (note that the calculator marks these on the screen in Figure 3(d)). However, the cursor already lies between these marks, so we typically just hit ENTER at this point. We suggest you do so also.

(a) (b) (c) (d) Figure 3. Using the zero utility to ﬁnd an x-intercept.

The calculator responds by marking the x-intercept and reporting its x-value at the bottom of the screen, as shown in Figure 4(a). This is one of the zeros of the function. Note that this value of −1.5 agrees nicely with our hand calculated result −3/2 in Example 6. We followed precisely the same procedure outlined above to ﬁnd the second x-intercept shown in Figure 4(b). Note that it also agrees with the hand calculated solution of Example 6.

In a similar vein, the point where the graph of a function crosses the y-axis is called the y-intercept of the graph of the function. In Figure 1 the y-intercept of the parabola is (0, −6). Note that the x-coordinate of this y-intercept is zero. Thus, the process for ﬁnding y-intercepts should be clear.

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(a) (b) Figure 4. The zeros of f(x) − 2x2 − 7x − 15.

Finding y-intercepts. To ﬁnd the y-intercepts of the graph of any function, set x = 0 and solve for y. Alternatively, if function notation is used, simply evaluate f(0).

I Example 9. Find the y-intercept of the quadratic function deﬁned by f(x) = x2 − 3x − 11.

Evaluate the function at x = 0.

f(0) = (0)2 − 3(0) − 11 = −11.

The coordinates of the y-intercept are (0, −11).

Putting it All Together

We will ﬁnd both x- and y-intercepts extremely useful when drawing the graph of a quadratic function.

2 I Example 10. Place the quadratic function y = x + 2x − 24 in vertex form. Plot the vertex and axis of symmetry and label them with their coordinates and equation, respectively. Find and plot the x- and y-intercepts of the parabola and label them with their coordinates.

Take half of the coeﬃcient of x, square, then add and subtract this amount to balance the equation. Factor and combine coeﬃcients. y = x2 + 2x + 1 − 1 − 24 y = (x + 1)2 − 25

The graph is a parabola that opens upward; it is shifted 1 unit to the left and 25 units downward. This information is enough to plot and label the vertex, then plot and label the axis of symmetry, as shown in Figure 5(a). To ﬁnd the x-intercepts, let y = 0 in y = x2 + 2x − 24.

0 = x2 + 2x − 24

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The leading coeﬃcient is a 1. The integer pair −4 and 6 has product −24 and sum 2. Thus, the right-hand side factors as follows.

0 = (x + 6)(x − 4)

In order that this product equals zero, either

x + 6 = 0 or x − 4 = 0.

Solve each of these linear equations independently.

x = −6 or x = 4.

Recall that we let y = 0. We’ve found two solutions, x = −6 and x = 4. Thus, we have x-intercepts at (−6, 0) and (4, 0), as pictured in Figure 5(b). Finally, to ﬁnd the y-intercept, let x = 0 in y = x2 +2x−24. With this substitution, y = −24. Thus, the y-intercept is (0, −24), as pictured in Figure 5(c). Note that we’ve also included the mirror image of the y-intercept across the axis of symmetry. y y y

x x x (−6, 0) (4, 0) (−6, 0) (4, 0)

(−1, −25) (−1, −25) (−1, −25) (0, −24)

x = −1 x = −1 x = −1 (a) Plotting the vertex (b) Adding the x-intercepts (c) Adding the and axis of symmetry. provides added accuracy. y-intercept and its mirror image provides an excellent ﬁnal graph. Figure 5.

Let’s look at one ﬁnal example.

2 I Example 11. Plot the parabola represented by the equation f(x) = −2x −7x+15. Plot and label the vertex, axis of symmetry, and the x- and y-intercepts.

First, factor out a −2. 7 15 f(x) = −2 x2 + x − 2 2

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Half3 of 7/2 is 7/4. Squared, this amounts to 49/16. Add and subtract this last amount to keep the equation balanced. 7 49 49 15 f(x) = −2 x2 + x + − − 2 16 16 2 The ﬁrst three terms inside the parentheses form a perfect square trinomial. The last two constants are combined with a common denominator. 7 49 49 120 f(x) = −2 x2 + x + − − 2 16 16 16 " # 72 169 f(x) = −2 x + − 4 16

Finally, redistribute the −2. 72 169 f(x) = −2 x + + 4 8 The graph of this last equation is a parabola that opens downward, translated 7/4 units to the left and 169/8 units upward. This is enough information to plot and label the vertex and axis of symmetry, as shown in Figure 6(a). To ﬁnd the y-intercepts, set f(x) = 0 in f(x) = −2x2 − 7x + 15. We will also multiply both sides of the resulting equation by −1. 0 = −2x2 − 7x + 15 0 = 2x2 + 7x − 15

After comparing 2x2 + 7x − 15 with ax2 + bx + c, we note that the integer pair −3 and 10 have product equal to ac = −30 and sum equal to b = 7. Use this pair to express the middle term of 2x2 + 7x − 15 as a sum and then factor by grouping. 0 = 2x2 − 3x + 10x − 15 0 = x(2x − 3) + 5(2x − 3) 0 = (x + 5)(2x − 3)

By the zero product property, either

x + 5 = 0 or 2x − 3 = 0.

Solve these linear equations independently. 3 x = −5 or x = 2 These x-values are the zeros of f (they make f(x) = 0), so we have x-intercepts at (−5, 0) and (3/2, 0), as shown in Figure 6(b).

3 1 7 7 2 · 2 = 4

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y y y

(−7/4, 169/8) (−7/4, 169/8) (−7/4, 169/8) (0, 15) (−5, 0) (3/2, 0) (−5, 0) (3/2, 0) x x x

x = −7/4 x = −7/4 x = −7/4 (a) Plotting the vertex (b) Adding the x-intercepts (c) Adding the y-intercept and axis of symmetry. provides added accuracy. and its mirror image provide an excellent ﬁnal graph. Figure 6.

Finally, to ﬁnd the y-intercept, set x = 0 in f(x) = −2x2 − 7x + 15 to get f(0) = 15. Note the positioning of the y-intercept (0, 15) and its mirror image across the axis of symmetry in Figure 6(c).

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