Chapter 4.1, Problem 78AYU

To determine

To analyze: the given polynomial

Answer to Problem 78AYU

  

  

Explanation of Solution

Given: $f\left(x\right)=x^2\left(x-3\right)\left(x+4\right)$

Calculation:

Given: $f\left(x\right)=x^2\left(x-3\right)\left(x+4\right)$

Step1: Determine the end behavior of the graph of the function

Expand the polynomial to write it in the form

  $f\left(x\right)=\text{}\text{}{a}_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$

  $f\left(x\right)=x^2\left(x-3\right)\left(x+4\right)$

  $=x^2\left(x^2+4x-3x-12\right)$ Multiply

  $=\left(x^4+4x^3-3x^3-12x^2\right)$

  $=x^4+x^3-12x^2$ Combine like terms

The polynomial function $f$ is of degree $4$ . The graph of $f$ behaves like $y=x^4$ for large of .

Step2: find the $x$ - and $y$ -intercepts of the graph of the function

The $y$ -intercepts is $f\left(0\right)=0$ . To find the $x$ -intercepts, we solve

  

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

So

  $x^2=0$ or $x-3=0$ or $x+4=0$

  $x=0$ or $x=3$ or $x=-4$

The $x$ -intercepts are $-4,0$ and 3.

Step3: Determine the zeros of the function and their multiplicity. Use this information to determine whether the graph crosses or touches the $x-axis$ at each $x$ -intercept.

The zeros of $f$ are $-4,0$ and 3. The zero $-4$ is a zero of multiplicity $1$ , so the graph of $f$ crosses the $x-axis$ at $x=-4$ . The zero $0$ is a zero of multiplicity $2$ , so the graph $f$ touches the $x-axis$ at $x=0$ . The zero $3$ is a zero of multiplicity $1$ , so the graph of $f$ crosses the $x-axis$ at $x=3$ .

Step4: Determine the maximum number of turning points on the graph of the function. Because the polynomial function is of degree $4$ (step1), the graph of the function will have atmost $4-1=3$ points.

Step5: Determine the behavior of the graph of $f$ near each $x$ -intercept

The three $x$ -intercept are $-4,0$ and $3$ .

Near

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

  $\approx {\left(-4\right)}^2\left(-4-3\right)\left(x+4\right)$

  $=16\left(-7\right)\left(x+4\right)$ A line with slope -112

  $=-112\left(x+4\right)$

  $=-112x-448$

Near $0$ : $f\left(x\right)=x^2\left(x-3\right)\left(x+4\right)$

  $\approx x^2\left(0-3\right)\left(0+4\right)$

  $=x^2\left(-12\right)$ A parabola that opens down

  $=-12x^2$

Near $3$ : $f\left(x\right)=x^2\left(x-3\right)\left(x+4\right)$

  $\approx {\left(3\right)}^2\left(x-3\right)\left(3+4\right)$

  $=9\left(7\right)\left(x-3\right)$ A line with slope 63

  $=63\left(x-3\right)$

  $=63x-189$

Step 6: Put all the information form Steps 1 through 5 together to obtain the graph of $f$

Figure 1(a) illustrates the information obtained from Steps 1 through 5. We evaluate $f$ at $-3$ , $2$ to help establish the scale on the $y-axis$ . The graph of $f$ is given in Figure 1(b).

Figure 1

  

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 5.