Chapter 2.2, Problem 14E

To determine

To identify (match):The graph of a polynomial from the given list of graphs.

Answer to Problem 14E

The polynomial function $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ represents the graph (d) from the list.

Explanation of Solution

Given information:

The given polynomial function is $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ .

Concept used:

We use leading coefficient test. When degree of polynomial n is odd and the leading coefficient is positive $a_n<0$ then the graph falls to the right and rises to the left.

The givenpolynomial function is $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ of odd degree.

Test-1: Since, the leading coefficient $a_n=-\frac{1}{3}<0$ the graph falls to the right and rises to the left.

Test-2: The x-intercepts of the polynomial function $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ are obtained.

  $f\left(x\right)=0\Rightarrow f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}=0$

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

  $\Rightarrow x^3-3x^2+4=0$

  $\Rightarrow x^3-3x^2+4=x^3+x^2-4x^2-4x+4x+4$

  $\Rightarrow x^2\left(x+1\right)-4x\left(x+1\right)+4\left(x+1\right)$

  $\Rightarrow \left(x+1\right)\left(x^2-4x+4\right)$

  $\Rightarrow \left(x+1\right){\left(x-2\right)}^2$...............(1)

The x- intercepts of the graph of $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ are $\left(-1,0\right)$ and $\left(2,0\right)$ .

Test-3: The y- intercept of the polynomial function $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ is $\left(0,f(0)\right)\equiv \left(0,-\frac{4}{3}\right)$

We see that the graph (d) from the list of polynomial qualify all three tests.

  

Hence, the polynomial function $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ represents the graph (d) from the list.

Conclusion:

The polynomial function $f\left(x\right)=-\frac{1}{3}{x}^3+x^2-\frac{4}{3}$ represents the graph (d) from the list.