Precalculus

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 5, Problem 10CR

(a)

To determine

The end behavior of polynomial $f(x)=3x4−15x3−12x2+60x$ .

(a)

Expert Solution

Answer to Problem 10CR

Solution:

The graph of the function $f(x)=3x4−15x3−12x2+60x$ behaves like $y=3x4$ for large value of $|x|$ .

Explanation of Solution

Given information:

The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

Explanation:

The leading term of $f(x)$ is $3x4$ .

The graph of the function $f(x)=3x4−15x3−12x2+60x$ behaves like $y=3x4$ for large value of $|x|$ .

A polynomial function $f(x)$ with even degree and positive leading coefficient has end behavior $x→−∞, f(x)→∞ and x→∞, f(x)→∞$ .

The degree of polynomial $f(x)=3x4−15x3−12x2+60x$ is $4$ , which is even, and the leading coefficient is $3$ , which is positive.

Therefore, the end behavior of polynomial $f(x)=3x4−15x3−12x2+60x$ would be $x→−∞, f(x)→∞ and x→∞, f(x)→∞$ .

(b)

To determine

The $x$ and $y$ intercepts of the graph of $f(x)=3x4−15x3−12x2+60x$ .

(b)

Expert Solution

Answer to Problem 10CR

Solution:

The $y−$ intercept of the graph of $f(x)=3x4−15x3−12x2+60x$ is $(0, 0)_$ , and $x−$ intercepts are $(−2, 0), (0, 0),(2, 0),  (5, 0)_$ .

Explanation of Solution

Given information:

The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

Explanation:

To find $y−$ intercept of the graph of $f(x)$ , substitute $x=0$ in $y=3x4−15x3−12x2+60x$ ,

$⇒ y=3(0)4−15(0)3−12(0)2+60(0)$

$⇒ y=0$

Therefore, the $y−$ intercept of the graph of $f(x)=3x4−15x3−12x2+60x$ is $(0, 0)$ .

To find $x-$ intercept of the graph of $f(x)$ , substitute $y=0$ in $y=3x4−15x3−12x2+60x$ ,

$⇒ 3x4−15x3−12x2+60x=0$

As $x$ is common factor, it can be factored out.

Using distributive property,

$⇒ 3x(x3−5x2−4x+20)=0$

$⇒ 3x=0, x3−5x2−4x+20=0$

$⇒ x=0, x3−5x2−4x+20=0$

The equation $x3−5x2−4x+20=0$ can be solved by factoring method.

Factor out $x2$ from first two terms and $−4$ from last two terms of equation $x3−5x2−4x+20=0$ .

Using distributive property,

$⇒x2(x−5)−4(x−5)=0$

$⇒(x2−4)(x−5)=0$

$⇒ x2−4=0, x−5=0$

Add $5$ to both sides of $x−5=0$ , and add $4$ to both sides of $x2−4=0$ ,

$⇒ x=5, x2=4$

Take square root on both sides of $x2=4$ ,

$x2=±4$

$⇒ x=−2, 2$

The $x−$ intercepts of function $f(x)$ are at $x=−2, 0, 2, 5$ .

Therefore, the $x−$ intercepts of function $f(x)$ are $(−2, 0), (0, 0),(2, 0),  (5, 0)$ .

(c)

To determine

The real zeros of polynomial $f(x)=3x4−15x3−12x2+60x$ with their multiplicities, and to determine if the graph crosses or touches the $x−$ axis at each intercept.

(c)

Expert Solution

Answer to Problem 10CR

Solution:

The real zeros of polynomial $f(x)=3x4−15x3−12x2+60x$ are $x=−2, 0, 2, 5_$ , with each zero havingmultiplicity $1$ , and the graph crosses $x−$ axis at each intercept.

Explanation of Solution

Given information:

The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

Explanation:

The real zeros of polynomial $f(x)=3x4−15x3−12x2+60x$ are nothing but the $x−$ intercepts of $f(x)=3x4−15x3−12x2+60x$ .

From part (b), $x−$ intercepts of $f(x)=3x4−15x3−12x2+60x$ are $x=−2, 0, 2, 5$ .

Therefore, the real zeros of $f(x)=3x4−15x3−12x2+60x$ are $x=−2, 0, 2, 5$ .

The degree of $f(x)$ is $4$ , and there are $4$ distinct real zeros of $f(x)$ . Since each zero appears once, the multiplicity of each real zero is $1$ .

The multiplicity of each interceptis $1$ , which is odd. So, graph of $f(x)$ will cross $x−$ axis at each intercept.

(d)

To determine

The maximum number of turning points on the graph of polynomial $f(x)=3x4−15x3−12x2+60x$ .

(d)

Expert Solution

Answer to Problem 10CR

Solution:

The graph of polynomial $f(x)=3x4−15x3−12x2+60x$ has at most $3$ turning points.

Explanation of Solution

Given information:

The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

Explanation:

If a polynomial function $f$ is of degree $n$ , then the graph of $f$ has at most $n−1$ turning points.

The degree of polynomial $f(x)=3x4−15x3−12x2+60x$ is $4$ . So, graph of $f(x)=3x4−15x3−12x2+60x$ will have at most $4−1=3$ number of turning points.

Therefore, the graph of polynomial $f(x)=3x4−15x3−12x2+60x$ has at most $3$ turning points.

(e)

To determine

To graph: The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

(e)

Expert Solution

Explanation of Solution

Given information:

The polynomial function $f(x)=3x4−15x3−12x2+60x$ .

Graph:

The graph of the function $f(x)=3x4−15x3−12x2+60x$ behaves like $y=3x4$ for large value of $|x|$ .

The $x−$ intercepts of polynomial $f(x)=3x4−15x3−12x2+60x$ are $(−2, 0), (0, 0),(2, 0),  (5, 0)$ , and the $y−$ intercept is $(0, 0)$ .

Create a table of values as shown below:

$xy(x, y)−1−54(−1, −54)136(1, 36)3−90(3, −90)4−144(4, −144)$

Plot the $x$ and $y$ intercepts and points shown in the table on $xy$ plane, and draw the curve passing through all these points.

Therefore, the graph of $f(x)=3x4−15x3−12x2+60x$ is:

Precalculus, Chapter 5, Problem 10CR

Interpretation:

The graph represents the polynomial function $f(x)=3x4−15x3−12x2+60x$ , which has real zeros $x=−2, 0, 2, 5$ , and behaves the same as $y=3x4$ for large value of $|x|$ .

Chapter 5, Problem 9CR

Chapter 5, Problem 11CR

Chapter 5 Solutions

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