Question

use a graphing calculator to graph ( f(x)=x^{3}+4x^{2}-2x - 20 ) and find the following features of the graph. round answers to the nearest hundredth, if necessary.
list the ( x )-intercept of the graph.
( x=)
find the intervals on which the function is increasing.
( x< ) and ( x> )
find the interval on which the function is decreasing.

Explanation:

Step1: Find the x - intercept

The x - intercepts of the graph of \(y = f(x)\) are the solutions of the equation \(f(x)=0\), i.e., \(x^{3}+4x^{2}-2x - 20=0\). Using a graphing calculator, we find that \(x = 2\) is an x - intercept.

Step2: Find the derivative of the function

The derivative of \(y=f(x)=x^{3}+4x^{2}-2x - 20\) is \(y'=f'(x)=3x^{2}+8x - 2\).

Step3: Find the critical points

Set \(y' = 0\), so \(3x^{2}+8x - 2=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 3\), \(b = 8\), \(c=-2\), we have \(x=\frac{-8\pm\sqrt{64+24}}{6}=\frac{-8\pm\sqrt{88}}{6}=\frac{-8\pm2\sqrt{22}}{6}=\frac{-4\pm\sqrt{22}}{3}\).
\(x_1=\frac{-4-\sqrt{22}}{3}\approx - 2.91\), \(x_2=\frac{-4 + \sqrt{22}}{3}\approx0.24\)

Step4: Determine the intervals of increase and decrease

We test the intervals \((-\infty,\frac{-4-\sqrt{22}}{3})\), \((\frac{-4-\sqrt{22}}{3},\frac{-4+\sqrt{22}}{3})\) and \((\frac{-4+\sqrt{22}}{3},\infty)\) using a test point in each interval.
For the interval \((-\infty,\frac{-4-\sqrt{22}}{3})\), let \(x=-3\). Then \(y'=3\times(-3)^{2}+8\times(-3)-2=27-24 - 2=1>0\)
For the interval \((\frac{-4-\sqrt{22}}{3},\frac{-4+\sqrt{22}}{3})\), let \(x = 0\). Then \(y'=3\times0^{2}+8\times0-2=-2<0\)
For the interval \((\frac{-4+\sqrt{22}}{3},\infty)\), let \(x = 1\). Then \(y'=3\times1^{2}+8\times1-2=3 + 8-2=9>0\)

Answer:

The x - intercept is \(x = 2\).
The function is increasing for \(x<-2.91\) and \(x>0.24\).
The function is decreasing for \(-2.91