Chapter 2.2, Problem 85E

To determine

To graph: The polynomial function $f\left(x\right)$ by (i) applying leading coefficient test, (ii) finding the zeros of the polynomial , (iii) plotting sufficient solution points , and (iv) drawing a continuous curve through the points.

Explanation of Solution

Given data:

The given polynomial function $f\left(x\right)=x^3-9x$ .

Concept used:

Leading coefficient test: If the polynomial function $f\left(x\right)$ is of odd degree and leading coefficient to is positive then graph of function falls to the left and rises to the right.

Zeros of thepolynomial are the points $\left(a,f(a)=0\right)$ .

Graph:

The given polynomial is $f\left(x\right)=x^3-9x$ .

(i) Since the leading coefficient of given function is positive, therefore, the graph of function falls to the left and rises to the right.

(ii) Since, $f\left(x\right)=x^3-9x=0\Rightarrow x\left(x^2-9\right)=0\Rightarrow x=0,\pm 3$ . Therefore, zeros of the polynomial are $\left(0,0\right)$ , $\left(-3,0\right)$ , and $\left(3,0\right).$

(iii) Some solutions of the function (computed by taking arbitrary values of x ) are :

x	- 3.5	-2	-1	1	2	3.5
$y=f\left(x\right)$	-11.375	10	8	-8	-10	11.375

(iv). The continuous curve joining the calculated points (zeros and solutions), the graph of the polynomial function $f\left(x\right)=x^3-9x$ is shown in Figure-1.

  

Interpretation of the Graph:

The function $f\left(x\right)$ intersects x -axis at points $\left(-3,0\right)$ , $\left(0,0\right)$ and $\left(3,0\right)$ , so points $x=-3$ , $x=0$ and $x=3$ are zeros of $f\left(x\right)$ . Since, function $f\left(x\right)$ is ofodd degree and leading coefficient is positive so graph of $f\left(x\right)$ falls to the left and rises to the right.