Chapter 2.2, Problem 13E

To determine

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

Answer to Problem 13E

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

Explanation of Solution

Given information:

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

Concept used:

We use leading coefficient test. When degree of polynomial n is even and the leading coefficient is negative then the graph falls to the left and right.

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

Test-1: Since, the leading coefficient $a_n=-\frac{1}{4}<0$ the graph falls to the left and right.

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

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

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

  $\Rightarrow x=0,x=\pm \sqrt{12}=\pm 2\sqrt{3}=\pm 3.464$...............(1)

The x- intercepts of the graph of $f\left(x\right)=-\frac{1}{4}{x}^4+3x^2$ are $\left(0,0\right)$ , and $\left(\pm 3.464,0\right)$ .

Test-3:Since $f\left(-x\right)=-\frac{1}{4}{\left(-x\right)}^4+3{\left(-x\right)}^2=-\frac{1}{4}{x}^4+3x^2=f\left(x\right)$ , the graph of $f\left(x\right)$ is symmetrical about y-axis.

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

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

  

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

Conclusion:

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