Each function in Figure 21 has even degree and is an even function exhib- iting symmetry about the y-axis. Each has domain 1-∞, ∞2 but restricted range 30, ∞2. These even functions are also continuous on their entire domain 1-∞, ∞2. However, they are decreasing on 1-∞, 02 and increasing on 10, ∞2, appearing as though they rise both to the left and to the right.
3.4 Polynomial Functions: Graphs, Applications, and Models
Graphs of f (x) = axn We can now graph polynomial functions of degree 3 or greater with real number coefficients and domains (because the graphs are in the real number plane). We begin by inspecting the graphs of several functions of the form
ƒ1x2 = axn, with a= 1.
The identity function ƒ1x2= x, the squaring function ƒ1x2= x2, and the cubing function ƒ1x2= x3 were graphed earlier using a general point-plotting method.
Each function in Figure 20 has odd degree and is an odd function exhibiting symmetry about the origin. Each has domain 1-∞, ∞2 and range 1-∞, ∞2 and is continuous on its entire domain 1-∞, ∞2. Additionally, these odd functions are increasing on their entire domain 1-∞, ∞2, appearing as though they fall to the left and rise to the right.
■ Graphs of f (x)=axn
■ Graphs of General Polynomial Functions
■ Behavior at Zeros
■ Turning Points and End Behavior
■ Graphing Techniques
■ Intermediate Value and Boundedness Theorems
■ Approximations of Real Zeros
■ Polynomial Models
x
y f(x) = x
0 1 –1–1
1 2
–2
–2 2
x y f(x) = x3
0 1 –1–1
1 2
–2
–2 2
x y f(x) = x5
Rises to the right
Falls to the left
0 1 –1–1
1 2
–2
–2 2
Figure 20
x y f(x) = x2
0 1 –1–1
1 2
–2
–2 2
x y f(x) = x4
0 1 –1–1
1 2
–2
–2 2
x y f(x) = x6
0 1 –1–1
1 2
–2
–2 2
Rises to the right Rises to
the left
Figure 21
The behaviors in the graphs of these basic polynomial functions as x increases (decreases) without bound also apply to more complicated polynomial functions.
Graphs of General Polynomial Functions As with quadratic functions, the absolute value of a in ƒ1x2 = axn determines the width of the graph.
• When a 71, the graph is stretched vertically, making it narrower.
• When 06 a 6 1, the graph is shrunk or compressed vertically, making it wider.
Compared to the graph of ƒ1x2 =axn, the following also hold true.
• The graph of ƒ1x2 = −axn is reflected across the x-axis.
• The graph of ƒ1x2 = axn+ k is translated (shifted) k units up if k70 and k units down if k60.
• The graph of ƒ1x2 = a1x− h2n is translated h units to the right if h70 and h units to the left if h60.
• The graph of ƒ1x2 = a1x− h2n+ k shows a combination of these transla- tions.
EXAMPLE 1 Examining Vertical and Horizontal Translations
Graph each polynomial function. Determine the largest open intervals of the domain over which each function is increasing or decreasing.
(a) ƒ1x2 =x5 - 2 (b) ƒ1x2= 1x +126 (c) ƒ1x2= -21x - 123 + 3 SOLUTION
(a) The graph of ƒ1x2= x5 - 2 will be the same as that of ƒ1x2 =x5, but trans- lated 2 units down. See Figure 22. This function is increasing on its entire domain 1-∞, ∞2.
(b) In ƒ1x2 = 1x+ 126, function ƒ has a graph like that of ƒ1x2 =x6, but because x + 1= x- 1-12,
it is translated 1 unit to the left. See Figure 23. This function is decreasing on 1-∞, -12 and increasing on 1-1, ∞2.
x y
f(x) = x5 – 2 0
2 4 6
–2–2 2
Figure 22
x y
f(x) = (x + 1)–2 –16 1 2 0 2 4 6
Figure 23
x y
f(x) = –2(x – 1)3 + 3
0 1 2 3
Figure 24 (c) The negative sign in -2 causes the graph of
ƒ1x2 = -21x- 123 + 3
to be reflected across the x-axis when compared with the graph of ƒ1x2 =x3. Because -2 71, the graph is stretched vertically when compared to the graph of ƒ1x2 = x3. As shown in Figure 24, the graph is also translated 1 unit to the right and 3 units up. This function is decreasing on its entire domain 1-∞, ∞2.
✔ Now Try Exercises 13, 15, and 19.
Unless otherwise restricted, the domain of a polynomial function is the set of all real numbers. Polynomial functions are smooth, continuous curves on the interval 1-∞, ∞2. The range of a polynomial function of odd degree is also the set of all real numbers.
367
3.4 Polynomial Functions: Graphs, Applications, and Models
The graphs in Figure 25 suggest that for every polynomial function ƒ of odd degree there is at least one real value of x that satisfies ƒ1x2 = 0. The real zeros correspond to the x-intercepts of the graph and can be determined by inspecting the factored form of each polynomial.
Odd Degree
x 0 1
–3 –2 –6
–12 6 y
f(x) = 2x3 + 8x2 + 2x – 12 f(x) = 2(x – 1)(x + 2)(x + 3) Three
real zeros
x 0 2
–4 6 2 y
f(x) = –x3 + 2x2 – x + 2 f(x) = –(x – 2)(x – i)(x + i)
One real zero
x 1
–2 0 –4
8
2 y
f(x) = x5 + 4x4 + x3 – 10x2 – 4x + 8 f(x) = (x – 1)2(x + 2)3
Two real zeros
(a) (b) (c)
Figure 25
A polynomial function of even degree has a range of the form 1-∞, k4 or 3k, ∞2, for some real number k. Figure 26 shows two typical graphs.
Behavior at Zeros Figure 26(b) shows a sixth-degree polynomial function with three distinct zeros, yet the behavior of the graph at each zero is different. This behavior depends on the multiplicity of the zero as determined by the exponent on the corresponding factor. The factored form of the polynomial function ƒ1x2 is
-1x +2211x+ 1221x -123.
• 1x+ 22 is a factor of multiplicity 1. Therefore, the graph crosses the x-axis at 1-2, 02.
• 1x+ 12 is a factor of multiplicity 2. Therefore, the graph is tangent to the x-axis at 1-1, 02. This means that it touches the x-axis, then turns and changes behavior from decreasing to increasing similar to that of the squaring function ƒ1x2 =x2 at its zero.
• 1x- 12 is a factor of multiplicity 3. Therefore, the graph crosses the x-axis and is tangent to the x-axis at 11, 02. This causes a change in concavity (that is, how the graph opens upward or downward) at this x-intercept with behav- ior similar to that of the cubing function ƒ1x2 =x3 at its zero.
Even Degree
1 2 x –2
4 y
f(x) = x4 – 5x2 + 4
f(x) = (x – 1)(x + 1)(x – 2)(x + 2) Four
real zeros
k –1
x 1
–1
–2 0
2 y
f(x) = –x6 – x5 + 4x4 + 2x3 – 5x2 – x + 2 f(x) = –(x + 2)(x + 1)2(x – 1)3
The graph touches the x-axis and then turns at (–1, 0).
The graph crosses the x-axis and changes shape at (1, 0).
The graph crosses the x-axis at (–2, 0).
k
(a) (b)
Figure 26
Figure 27 generalizes the behavior of such graphs at their zeros.
c x c x c x
c x c x
c x
The graph crosses the x-axis at 1c, 02 if c is a zero of multiplicity 1.
The graph is tangent to the x-axis at 1c, 02 if c is a zero of even multiplicity.
The graph bounces, or turns, at c.
The graph crosses and is tangent to the x-axis at 1c, 02 if c is a zero of odd multiplicity greater than 1. The graph wiggles at c.
Figure 27
Turning Points and End Behavior The graphs in Figures 25 and 26
show that polynomial functions often have turning points where the function changes from increasing to decreasing or from decreasing to increasing.
Turning Points
A polynomial function of degree n has at most n- 1 turning points, with at least one turning point between each pair of successive zeros.
The end behavior of a polynomial graph is determined by the dominating term—that is, the term of greatest degree. A polynomial of the form
ƒ1x2 = anxn + an-1xn-1 +g+ a0 has the same end behavior as ƒ1x2= anxn. For example,
ƒ1x2= 2x3 +8x2 + 2x- 12
has the same end behavior as ƒ1x2= 2x3. It is large and positive for large posi- tive values of x, while it is large and negative for negative values of x with large absolute value. That is, it rises to the right and falls to the left.
Figure 25(a) shows that as x increases without bound, y does also. For the same graph, as x decreases without bound, y does also.
As xS∞, yS∞ and as xS -∞, yS -∞.
As x –∞, y ∞ (rises to the left)
As x ∞, y –∞
(falls to the right)
x 0 2
–4 6 2 y
f(x) = –x3 + 2x2 – x + 2 f(x) = –(x – 2)(x – i)(x + i) Figure 25(b) (repeated) x
0 1 –3 –2
–6
–12 6 y
f(x) = 2x3 + 8x2 + 2x – 12 f(x) = 2(x – 1)(x + 2)(x + 3)
As x ∞, y ∞ (rises to the right)
As x –∞, y –∞
(falls to the left)
Figure 25(a) (repeated)
The graph in Figure 25(b) has the same end behavior as ƒ1x2 = -x3. As xS∞, yS -∞ and as xS -∞, yS∞.
The graph of a polynomial function with a dominating term of even degree will show end behavior in the same direction. See Figure 26.
LOOKING AHEAD TO CALCULUS To find the x-coordinates of the two turning points of the graph of
ƒ1x2=2x3+8x2+2x-12, we can use the “maximum” and
“minimum” capabilities of a graphing calculator and determine that, to the nearest thousandth, they are -0.131 and
-2.535. In calculus, their exact values can be found by determining the zeros of the derivative function of ƒ1x2,
ƒ′1x2=6x2+16x+2, because the turning points occur precisely where the tangent line has slope 0. Using the quadratic formula would show that the zeros are
-4{213
3 ,
which agree with the calculator approximations.
369
3.4 Polynomial Functions: Graphs, Applications, and Models
End Behavior of Graphs of Polynomial Functions
Suppose that axn is the dominating term of a polynomial function ƒ of odd degree.
1. If a 7 0, then as xS∞, ƒ1x2S∞, and as xS -∞, ƒ1x2S -∞. There- fore, the end behavior of the graph is of the type shown in Figure 28(a). We symbolize it as .
2. If a60, then as xS∞, ƒ1x2S -∞, and as xS -∞, ƒ1x2S∞. There- fore, the end behavior of the graph is of the type shown in Figure 28(b). We symbolize it as .
a > 0 n odd
a < 0 n odd
(a) (b)
Figure 28
a > 0
n even a < 0 n even
(a) (b)
Figure 29
Suppose that axn is the dominating term of a polynomial function ƒ of even degree.
1. If a7 0, then as x S∞, ƒ1x2S∞. Therefore, the end behavior of the graph is of the type shown in Figure 29(a). We symbolize it as . 2. If a60, then as x S∞, ƒ1x2S -∞. Therefore, the end behavior of
the graph is of the type shown in Figure 29(b). We symbolize it as .
EXAMPLE 2 Determining End Behavior
The graphs of the polynomial functions defined as follows are shown in A–D.
ƒ1x2= x4 - x2+ 5x- 4, g1x2 = -x6 + x2 - 3x -4, h1x2= 3x3 - x2+ 2x- 4, and k1x2 = -x7 + x- 4
Based on the discussion of end behavior, match each function with its graph.
A. B. C. D.
SOLUTION
• Function ƒ has even degree and a dominating term with positive leading coef- ficient, as in C.
• Function g has even degree and a dominating term with negative leading coefficient, as in A.
• Function h has odd degree and a dominating term with positive coefficient, as in B.
• Function k has odd degree and a dominating term with negative coefficient, as in D.
✔ Now Try Exercises 21, 23, 25, and 27.
–2 x
y 30
0 2
–60
–2 x
y
30 0
2
–30 –20
x y
50
–2 2
0
–40
x y
40
–2 2
0
Graphing Techniques We have discussed several characteristics of the graphs of polynomial functions that are useful for graphing the function by hand.
A comprehensive graph of a polynomial function ƒ1x2 will show the following characteristics.
• all x-intercepts (indicating the real zeros) and the behavior of the graph at these zeros
• the y-intercept
• the sign of ƒ1x2 within the intervals formed by the x-intercepts
• enough of the domain to show the end behavior
In Examples 3 and 4, we sketch the graphs of two polynomial functions by hand. We use the following general guidelines.
Graphing a Polynomial Function
Let ƒ1x2= anxn+ an-1xn-1 + g+ a1x + a0, with an ≠0, be a polyno- mial function of degree n. To sketch its graph, follow these steps.
Step 1 Find the real zeros of ƒ. Plot the corresponding x-intercepts.
Step 2 Find ƒ102= a0. Plot the corresponding y-intercept.
Step 3 Use end behavior, whether the graph crosses, bounces on, or wiggles through the x-axis at the x-intercepts, and selected points as neces- sary to complete the graph.
EXAMPLE 3 Graphing a Polynomial Function Graph ƒ1x2 = 2x3 + 5x2- x - 6.
SOLUTION
Step 1 The possible rational zeros are {1, {2, {3, {6, {12 , and {32 . Use synthetic division to show that 1 is a zero.
1)2 5 -1 -6 2 7 6
2 7 6 0 ƒ112=0
We use the results of the synthetic division to factor as follows.
ƒ1x2 =1x - 1212x2 + 7x+ 62
ƒ1x2 =1x - 1212x+ 321x+ 22 Factor again.
Set each linear factor equal to 0, and then solve for x to find zeros. The three zeros of ƒ are 1, -32 , and -2. Plot the corresponding x-intercepts.
See Figure 30.
Step 2 ƒ102 = -6, so plot 10, -62. See Figure 30.
Step 3 The dominating term of ƒ1x2 is 2x3, so the graph will have end behavior similar to that of ƒ1x2 =x3. It will rise to the right and fall to the left as . See Figure 30. Each zero of ƒ1x2 occurs with multiplicity 1, meaning that the graph of ƒ1x2 will cross the x-axis at each of its zeros. Because the graph of a polynomial function has no breaks, gaps, or sudden jumps, we now have sufficient information to sketch the graph of ƒ1x2.
Rises to the right
Falls to the left
(–2, 0)
(1, 0) –3
2
x y
6
(0, –6) Q , 0R0
Figure 30
371
3.4 Polynomial Functions: Graphs, Applications, and Models
Begin sketching at either end of the graph with the appropriate end behavior, and draw a smooth curve that crosses the x-axis at each zero, has a turning point between successive zeros, and passes through the y-intercept as shown in Figure 31.
Additional points may be used to verify whether the graph is above or below the x-axis between the zeros and to add detail to the sketch of the graph. The zeros divide the x-axis into four intervals:
1-∞, -22, a-2, -3
2b, a-3
2 , 1b , and 11, ∞2.
Select an x-value as a test point in each interval, and substitute it into the equation for ƒ1x2 to determine additional points on the graph. A typical selection of test points and the results of the tests are shown in the table.
Interval
Test Point x
Value of ƒ1x2
Sign of ƒ1x2
Graph Above or Below x-Axis 1-∞, -22 -3 -12 Negative Below
A-2, -32B -74 11
32 Positive Above
A-32 , 1B 0 -6 Negative Below
11, ∞2 2 28 Positive Above
✔ Now Try Exercise 29.
x y
6
–6
f (x) = 2x3 + 5x2 – x – 6 f (x) = (x – 1)(2x + 3)(x + 2)
0 1 2
–2 –1
Figure 31
EXAMPLE 4 Graphing a Polynomial Function Graph ƒ1x2 = -1x - 121x - 321x+ 222.
SOLUTION
Step 1 Because the polynomial is given in factored form, the zeros can be de- termined by inspection. They are 1, 3, and -2. Plot the corresponding x-intercepts of the graph of ƒ1x2. See Figure 32.
Step 2 ƒ102 = -10 -1210 - 3210 + 222 Find ƒ102.
ƒ102 = -1-121-321222 Simplify in parentheses.
ƒ102 = -12 The y-intercept is 10, -122. Plot the y-intercept 10, -122. See Figure 32.
Step 3 The dominating term of ƒ1x2 can be found by multiplying the factors and identifying the term of greatest degree. Here it is -1x21x21x22= -x4, indicating that the end behavior of the graph is . Because 1 and 3 are zeros of multiplicity 1, the graph will cross the x-axis at these zeros. The graph of ƒ1x2 will touch the x-axis at -2 and then turn and change direc- tion because it is a zero of even multiplicity.
Begin at either end of the graph with the appropriate end behavior and draw a smooth curve that crosses the x-axis at 1 and 3 and that touches the x-axis at -2, then turns and changes direction. The graph will also pass through the y-intercept 10, -122. See Figure 32.
Using test points within intervals formed by the x-intercepts is a good way to add detail to the graph and verify the accuracy of the sketch.
A typical selection of test points is 1-3, -242, 1-1, -82, 12, 162, and
14, -1082. ✔ Now Try Exercise 33.
x 1 3 –2 0
–6 –12 12 6 y
f(x) = –(x – 1)(x – 3)(x + 2)2 Figure 32
We emphasize the important relationships among the following concepts.
• the x-intercepts of the graph of y = ƒ1x2
• the zeros of the function ƒ
• the solutions of the equation ƒ1x2= 0
• the factors of ƒ1x2
For example, the graph of the function
ƒ1x2 =2x3 + 5x2 - x- 6 Example 3 ƒ1x2 =1x - 1212x+ 321x+ 22 Factored form
has x-intercepts 11, 02, A-32 , 0B, and 1-2, 02 as shown in Figure 31 on the previ- ous page. Because 1, -32 , and -2 are the x-values where the function is 0, they are the zeros of ƒ. Also, 1, -32 , and -2 are the solutions of the polynomial equation
2x3 +5x2 - x- 6= 0.
This discussion is summarized as follows.
x 1 3 –2 0
–6 –12 12 6 y Graph bounces at x = –2.
Graph crosses at x = 1, 3.
Figure 32 (repeated)
NOTE It is possible to reverse the process of Example 4 and write the polynomial function from its graph if the zeros and any other point on the graph are known. Suppose that we are asked to find a polynomial function of least degree having the graph shown in Figure 32 (repeated in the mar- gin). Because the graph crosses the x-axis at 1 and 3 and bounces at -2, we know that the factored form of the function is as follows.
Multiplicity Multiplicity one two ƒ1x2= a1x- 1211x - 3211x+ 222
Now find the value of a by substituting the x- and y-values of any other point on the graph, say 10, -122, into this function and solving for a.
ƒ1x2= a1x- 121x -321x + 222
-12= a10- 1210- 3210+ 222 Let x=0 and y= -12.
-12= a1122 Simplify.
a= -1 Divide by 12. Interchange sides.
Verify in Example 4 that the polynomial function is ƒ1x2= -1x - 121x- 321x+ 222.
Exercises of this type are labeled Connecting Graphs with Equations.
Relationships among x-Intercepts, Zeros, Solutions, and Factors If ƒ is a polynomial function and 1c, 02 is an x-intercept of the graph of y =ƒ1x2, then
c is a zero of ƒ, c is a solution of ƒ1x2 = 0, and x− c is a factor of ƒ1x2.
373
3.4 Polynomial Functions: Graphs, Applications, and Models
Intermediate Value and Boundedness Theorems As Examples 3 and 4 show, one key to graphing a polynomial function is locating its zeros. In the special case where the potential zeros are rational numbers, the zeros are found by the rational zeros theorem.
Occasionally, irrational zeros can be found by inspection. For instance, ƒ1x2= x3 -2 has the irrational zero 232.
The next two theorems apply to the zeros of every polynomial function with real coefficients. The first theorem uses the fact that graphs of polynomial func- tions are continuous curves. The proof requires advanced methods, so it is not given here. Figure 33 illustrates the theorem.
x y
f(a)
f(c) = 0 f(b) a c b
y = f(x)
Figure 33
Intermediate Value Theorem
If ƒ1x2 is a polynomial function with only real coefficients, and if for real numbers a and b the values ƒ1a2 and ƒ1b2 are opposite in sign, then there exists at least one real zero between a and b.
This theorem helps identify intervals where zeros of polynomial functions are located. If ƒ1a2 and ƒ1b2 are opposite in sign, then 0 is between ƒ1a2 and ƒ1b2, and so there must be a number c between a and b where ƒ1c2 =0.
CAUTION Be careful when interpreting the intermediate value theorem.
If ƒ1a2 and ƒ1b2 are not opposite in sign, it does not necessarily mean that there is no zero between a and b. In Figure 35, ƒ1a2 and ƒ1b2 are both nega- tive, but -3 and -1, which are between a and b, are zeros of ƒ1x2.
GRAPHING CALCULATOR SOLUTION
The graphing calculator screen in Figure 34 indi- cates that this zero is approximately 2.2469796.
(Notice that there are two other zeros as well.) ALGEBRAIC SOLUTION
Use synthetic division to find ƒ122 and ƒ132. 2)1 -2 -1 1
2 0 -2
1 0 -1 -1= ƒ122 3)1 -2 -1 1
3 3 6 1 1 2 7 = ƒ132
Because ƒ122 is negative and ƒ132 is positive, by the intermediate value theorem there must be a real zero between 2 and 3.
−3.1
−4.7
3.1
4.7
Figure 34
✔ Now Try Exercise 49.
EXAMPLE 5 Locating a Zero
Use synthetic division and a graph to show that ƒ1x2= x3 - 2x2 - x +1 has a real zero between 2 and 3.
The intermediate value theorem for polynomials helps limit the search for real zeros to smaller and smaller intervals. In Example 5, we used the theo- rem to verify that there is a real zero between 2 and 3. To locate the zero more accurately, we can use the theorem repeatedly. (Prior to modern-day methods involving calculators and computers, this was done by hand.)
x y
a f(b)
f(b) < 0 f(a) < 0 f(a)
b y = f(x) –3 –10
Figure 35
The boundedness theorem shows how the bottom row of a synthetic division is used to place upper and lower bounds on possible real zeros of a polynomial function.
Boundedness Theorem
Let ƒ1x2 be a polynomial function of degree n Ú 1 with real coefficients and with a positive leading coefficient. Suppose ƒ1x2 is divided syntheti- cally by x- c.
(a) If c7 0 and all numbers in the bottom row of the synthetic division are nonnegative, then ƒ1x2 has no zero greater than c.
(b) If c6 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then ƒ1x2 has no zero less than c.
Proof We outline the proof of part (a). The proof for part (b) is similar.
By the division algorithm, if ƒ1x2 is divided by x - c, then for some q1x2 and r,
ƒ1x2 = 1x- c2q1x2 +r,
where all coefficients of q1x2 are nonnegative, rÚ 0, and c70. If x7c, then x- c70. Because q1x270 and r Ú 0,
ƒ1x2= 1x- c2q1x2 +r7 0.
This means that ƒ1x2 will never be 0 for x7 c.
EXAMPLE 6 Using the Boundedness Theorem
Show that the real zeros of ƒ1x2 = 2x4 - 5x3 + 3x+ 1 satisfy these conditions.
(a) No real zero is greater than 3. (b) No real zero is less than -1.
SOLUTION
(a) Because ƒ1x2 has real coefficients and the leading coefficient, 2, is positive, use the boundedness theorem. Divide ƒ1x2 synthetically by x-3.
3)2 -5 0 3 1 6 3 9 36
2 1 3 12 37 All are nonnegative.
Here 370 and all numbers in the last row of the synthetic division are non- negative, so ƒ1x2 has no real zero greater than 3.
(b) We use the boundedness theorem again and divide ƒ1x2 synthetically by x- 1-12, or x+ 1.
-1)2 -5 0 3 1 -2 7 -7 4
2 -7 7 -4 5 These numbers alternate in sign.
Here -160 and the numbers in the last row alternate in sign, so ƒ1x2 has no real zero less than -1.
✔ Now Try Exercises 57 and 59.
375
3.4 Polynomial Functions: Graphs, Applications, and Models
Approximations of Real Zeros We can approximate the irrational real zeros of a polynomial function using a graphing calculator.
EXAMPLE 7 Approximating Real Zeros of a Polynomial Function Approximate the real zeros of ƒ1x2= x4 -6x3 + 8x2 + 2x- 1.
SOLUTION The dominating term is x4, so the graph will have end behavior similar to the graph of ƒ1x2= x4, which is positive for all values of x with large absolute values. That is, the end behavior is up at the left and the right, . There are at most four real zeros because the polynomial is fourth-degree.
Since ƒ102 = -1, the y-intercept is 10, -12. Because the end behavior is positive on the left and the right, by the intermediate value theorem ƒ has at least one real zero on either side of x= 0. To approximate the zeros, we use a graph- ing calculator. The graph in Figure 36 shows that there are four real zeros, and the table indicates that they are between
-1 and 0, 0 and 1, 2 and 3, and 3 and 4 because there is a sign change in ƒ1x2 =y1 in each case.
Using a calculator, we can find zeros to a great degree of accuracy.
Figure 37 shows that the negative zero is approximately -0.4142136. Similarly, we find that the other three zeros are approximately
0.26794919, 2.4142136, and 3.7320508.
✔ Now Try Exercise 77.
−6.2
−4.7
6.2
4.7
Figure 36
−6.2
−4.7
6.2
4.7
Figure 37
Polynomial Models
EXAMPLE 8 Examining a Polynomial Model
The table shows the number of transactions, in millions, by users of bank debit cards for selected years.
(a) Using x= 0 to represent 1995, x= 3 to represent 1998, and so on, use the regression feature of a cal- culator to determine the quadratic function that best fits the data. Plot the data and the graph.
(b) Repeat part (a) for a cubic function (degree 3).
(c) Repeat part (a) for a quartic function (degree 4).
(d) The correlation coefficient, R, is a measure of the strength of the relationship between two variables.
The values of R and R2 are used to determine how well a regression model fits a set of data. The closer the value of R2 is to 1, the better the fit. Compare R2 for the three functions found in parts (a)–(c) to decide which function best fits the data.
Source: Statistical Abstract of the United States.
Year
Transactions (in millions)
1995 829
1998 3765
2000 5290
2004 14,106
2008 28,464
2012 44,351