Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Question

Chapter 4, Problem 11RE

To determine

To analyze : The polynomial function $f(x)=(x−1)2(x+3)(x+1)$ .

Expert Solution & Answer

Answer to Problem 11RE

The polynomial function $f(x)=x4+2x3−4x2−2x+3$ .

The y -intercept is 3.

The x -intercept is zero 1, -1 and -3.

Explanation of Solution

Given information:

$f(x)=(x−1)2(x+3)(x+1)$ .

Formula used:

The graph of a polynomial function $f(x)=anxn+an−1xn−1+...+a1x+a0;an≠0$

Degree of a polynomial function $f:n$

Maximum number of turning points : $n−1$

At a zero even multiplicity: The graph of f touches the x -axis.

At a zero odd multiplicity : The graph of f crosses the x -axis.

For large $|x|$ , the graph of f behaves like the graph of $y=anxn$ .

Calculation:

Consider ,

$f(x)=(x−1)2(x+3)(x+1)$

Step 1.

Re-writing the polynomial is,

$=f(x)=(x−1)2(x+3)(x+1)$

$=x2[x2+4x+3]−2x[x2+4x+3]+1[x2+4x+3]=x4+4x3+3x2−2x3−8x2−6x+x2+4x+3=x4+2x3−4x2−2x+3$

The polynomial function f is of degree 4. The graph of f behaves like $x4$ .

Step 2.

The y −intercept :-

$=f(x)=x4+2x3−4x2−2x+3=f(0)=(0)+2(0)−4(0)−2(0)+3=f(0)=3$

The x -intercept :-

$=f(x)=0=x4+2x3−4x2−2x+3=0=(x−1)2(x+3)(x+1)=0$

$(x−1)2=0(x−1)(x−1)=0x=1,1$ or $x+1=0x=−1$ or $x+3=0x=−3$

Step3.

The zeros of f are 1, -1 and -3. The zero 1 is a zero of multiplicity 3, so the graph of f crosses the x - axis at $x=1$ . The zero -1 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=−1$ . The zero -3 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=−3$ .

Step4.

Since the degree of polynomial function is 4. Therefore, the graph of the function will have at most $=4−1=3$ turning points.

Step5.

The x -intercept are 1, -1 and -3.

The behavior of the graph of f near each x -intercept are as follows:-

Near $x=1$ ;

$f(x)=x4+2x3−4x2−2x+3$

$=(x−1)2(x+3)(x+1)=(x−1)2(1+3)(1+1)=(x−1)2(4)(2)=8(x−1)2$

A parabola that opens up .

Near $x=−1$

$f(x)=x4+2x3−4x2−2x+3$

$=(x−1)2(x+3)(x+1)=(−1−1)2(−1+3)(x+1)=(−2)2(2)(x+1)=(4)(2)(x+1)=8(x+1)=8x+8$

The line with slope 8.

Near $x=−3$

$f(x)=x4+2x3−4x2−2x+3$

$=(x−1)2(x+3)(x+1)=(−3−1)2(x+3)(−3+1)=(−4)2(x+3)(−2)=(16)(−2)(x+3)=−32(x+3)=−32x−96$

The line with slope -32.

Step6.

Using Step1- Step5 graph is drawn showing the x - intercept and y -intercept. With the help of graph the behavior of each x −intercept can be observed.

Precalculus Enhanced with Graphing Utilities, Chapter 4, Problem 11RE

Figure1.

Hence , the polynomial function f has been analyzed.

Chapter 4, Problem 10RE

Chapter 4, Problem 12RE

Chapter 4 Solutions

Precalculus Enhanced with Graphing Utilities

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The polynomial function f ( x ) = ( x − 1 ) 2 ( x + 3 ) ( x + 1 ) .

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Answer to Problem 11RE

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