Chapter 5, Problem 10CR

(a)

To determine

The end behavior of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Answer to Problem 10CR

Solution:

The graph of the function $f\left(x\right)=3x^4-15x^3-12x^2+60x$ behaves like $y=3x^4$ for large value of $\left|x\right|$.

Explanation of Solution

Given information:

The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Explanation:

The leading term of $f\left(x\right)$ is $3x^4$.

The graph of the function $f\left(x\right)=3x^4-15x^3-12x^2+60x$ behaves like $y=3x^4$ for large value of $\left|x\right|$.

A polynomial function $f\left(x\right)$ with even degree and positive leading coefficient has end behavior $x\to -\infty ,f\left(x\right)\to \infty \text{and }x\to \infty ,f\left(x\right)\to \infty$.

The degree of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ is $4$, which is even, and the leading coefficient is $3$, which is positive.

Therefore, the end behavior of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ would be $x\to -\infty ,f\left(x\right)\to \infty \text{and }x\to \infty ,f\left(x\right)\to \infty$.

(b)

To determine

The $x$ and $y$ intercepts of the graph of $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Answer to Problem 10CR

Solution:

The $y-$ intercept of the graph of $f\left(x\right)=3x^4-15x^3-12x^2+60x$ is $\underset{\_}{\left(0,0\right)}$, and $x-$ intercepts are $\underset{\_}{\left(-2,0\right),\left(0,0\right),\left(2,0\right),\left(5,0\right)}$.

Explanation of Solution

Given information:

The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Explanation:

To find $y-$ intercept of the graph of $f\left(x\right)$, substitute $x=0$ in $y=3x^4-15x^3-12x^2+60x$,

$\Rightarrow y=3{\left(0\right)}^4-15{\left(0\right)}^3-12{\left(0\right)}^2+60\left(0\right)$

$\Rightarrow y=0$

Therefore, the $y-$ intercept of the graph of $f\left(x\right)=3x^4-15x^3-12x^2+60x$ is $\left(0,0\right)$.

To find $x\text{-}$ intercept of the graph of $f\left(x\right)$, substitute $y=0$ in $y=3x^4-15x^3-12x^2+60x$,

$\Rightarrow 3x^4-15x^3-12x^2+60x=0$

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

Using distributive property,

$\Rightarrow 3x\left(x^3-5x^2-4x+20\right)=0$

$\Rightarrow 3x=0,x^3-5x^2-4x+20=0$

$\Rightarrow x=0,x^3-5x^2-4x+20=0$

The equation $x^3-5x^2-4x+20=0$ can be solved by factoring method.

Factor out $x^2$ from first two terms and $-4$ from last two terms of equation $x^3-5x^2-4x+20=0$.

Using distributive property,

$\Rightarrow x^2\left(x-5\right)-4\left(x-5\right)=0$

$\Rightarrow \left(x^2-4\right)\left(x-5\right)=0$

$\Rightarrow x^2-4=0,x-5=0$

Add $5$ to both sides of $x-5=0$, and add $4$ to both sides of $x^2-4=0$,

$\Rightarrow x=5,x^2=4$

Take square root on both sides of $x^2=4$,

$\sqrt{x^2}=\pm \sqrt{4}$

$\Rightarrow x=-2,2$

The $x-$ intercepts of function $f\left(x\right)$ are at $x=-2,0,2,5$.

Therefore, the $x-$ intercepts of function $f\left(x\right)$ are $\left(-2,0\right),\left(0,0\right),\left(2,0\right),\left(5,0\right)$.

(c)

To determine

The real zeros of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ with their multiplicities, and to determine if the graph crosses or touches the $x-$ axis at each intercept.

Answer to Problem 10CR

Solution:

The real zeros of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ are $\underset{\_}{x=-2,0,2,5}$, with each zero havingmultiplicity $1$, and the graph crosses $x-$ axis at each intercept.

Explanation of Solution

Given information:

The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Explanation:

The real zeros of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ are nothing but the $x-$ intercepts of $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

From part (b), $x-$ intercepts of $f\left(x\right)=3x^4-15x^3-12x^2+60x$ are $x=-2,0,2,5$.

Therefore, the real zeros of $f\left(x\right)=3x^4-15x^3-12x^2+60x$ are $x=-2,0,2,5$.

The degree of $f\left(x\right)$ is $4$, and there are $4$ distinct real zeros of $f\left(x\right)$. 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 $f\left(x\right)$ will cross $x-$ axis at each intercept.

(d)

To determine

The maximum number of turning points on the graph of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Answer to Problem 10CR

Solution:

The graph of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ has at most $3$ turning points.

Explanation of Solution

Given information:

The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

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 $f\left(x\right)=3x^4-15x^3-12x^2+60x$ is $4$. So, graph of $f\left(x\right)=3x^4-15x^3-12x^2+60x$ will have at most $4-1=3$ number of turning points.

Therefore, the graph of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ has at most $3$ turning points.

(e)

To determine

To graph: The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Explanation of Solution

Given information:

The polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$.

Graph:

The graph of the function $f\left(x\right)=3x^4-15x^3-12x^2+60x$ behaves like $y=3x^4$ for large value of $\left|x\right|$.

The $x-$ intercepts of polynomial $f\left(x\right)=3x^4-15x^3-12x^2+60x$ are $\left(-2,0\right),\left(0,0\right),\left(2,0\right),\left(5,0\right)$, and the $y-$ intercept is $\left(0,0\right)$.

Create a table of values as shown below:

$\begin{array}{ccc}x& y& \left(x,y\right)\\ -1& -54& \left(-1,-54\right)\\ 1& 36& \left(1,36\right)\\ 3& -90& \left(3,-90\right)\\ 4& -144& \left(4,-144\right)\end{array}$

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 $f\left(x\right)=3x^4-15x^3-12x^2+60x$ is:

Interpretation:

The graph represents the polynomial function $f\left(x\right)=3x^4-15x^3-12x^2+60x$, which has real zeros $x=-2,0,2,5$, and behaves the same as $y=3x^4$ for large value of $\left|x\right|$.