Chapter 2.2, Problem 89E

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 $f\left(x\right)=-x^4+9x^2-20$ .

Concept used:

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

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

Graph:

The given polynomial is $f\left(x\right)=-x^4+9x^2-20$ .

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

(ii) Since,function $f\left(x\right)=-x^4+9x^2-20=-\left(x^4-9x^2+20\right)$

  $=-\left(x^4-5x^2-4x^2+20\right)$

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

  $=-\left(x^2-4\right)(x^2-5)$ .Therefore, zeros of the polynomial $f\left(x\right)$ are $x=\pm 2,\pm \sqrt{5}$ ach of multiplicity 1.

The graph interests x -axis at points $\left(-2,0\right)$ , $\left(2,0\right)$ , $\left(-\sqrt{5},0\right)$ and $\left(\sqrt{5},0\right)$ .

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

x	-1	1	-3	3	-1.5	1.5
$y=f\left(x\right)$	-12	-12	-20	-20	-48.1	-48.1

(iv)The continuous curve joining the calculated points (zeros and solutions), the graph of the polynomial function $f\left(x\right)=-x^4+9x^2-20$ is shown in Figure-1.

  

Interpretation of the Graph:

The function $f\left(x\right)$ intersects x -axis at points $\left(\pm 2,0\right)$ and $\left(\pm \sqrt{5},0\right)$ , so $x=\pm 2$ and $x=\pm \sqrt{5}$ are zeros of $f\left(x\right)$ . Since, function $f\left(x\right)$ is ofeven degree and leading coefficient is negativeso graph of $f\left(x\right)$ falls to the left and to the right.