Q. 76 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75–80, for the given functions f and g, 

(a) Graph $f$ and $g$ on the same Cartesian plane.

(b) Solve $f\left(x\right)=g\left(x\right).$

 (c) Use the result of part (b) to label the points of intersection of the graphs of $f$ and $g.$

 (d) Shade the region for which $f\left(x\right)>g\left(x\right).$ that is, the region below f and above $g$. 

$f\left(x\right)=-2x-1; g\left(x\right)=x^2-9.$

Step-by-Step Solution

Answer

(a) The graph of $f$ and $g$ is

(b) The value of $x=-2,4.$

(c) The point of intersection is $\left(2,-5\right),\left(-4,7\right).$

(d) The shaded region is 

Part (a) Step 1. Given information

It is given that $f\left(x\right)=-2x-1; g\left(x\right)=x^2-9.$ We need to graph $f$ and $g$ on the same cartesian plane.

Step 2. Graph f x and g x

Part (b) Step 1. Given information

It is given that $f\left(x\right)=-2x-1; g\left(x\right)=x^2-9.$ We need to solve $f\left(x\right)=g\left(x\right).$

Step 2. Simplify

$f\left(x\right)=g\left(x\right).$

$-2x-1=x^2-9.$

$x^2+2x-8=0.$

$x=\frac{-2\pm \sqrt{2^2-4·1\left(-8\right)}}{2·1}.$

  $=\frac{-2\pm \sqrt{4+32}}{2·1}.$

  $=\frac{-2\pm \sqrt{36}}{2}.$

  $=\frac{-2\pm 6}{2}.$

  $x=\frac{-2+6}{2}=2, and x=\frac{-2-6}{2}=-4.$

The value of the function $f$ and $g$ are equal when $x=2,-4.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=-2x-1; g\left(x\right)=x^2-9.$ We need to use the result of part (b) to label the point of intersection.

Step 2. Simplify

Let $y=-2x-1.$

$y=g\left(x\right)=x^2-9.$

When $x=2, y=-2\left(2\right)-1=-5.$

When $x=-4, y=-2·\left(-4\right)-1=7.$

The point of intersection is $\left(2,-5\right),\left(-4,7\right).$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=-2x-1; g\left(x\right)=x^2-9.$ We need to label the shaded region for which $f\left(x\right)>g\left(x\right).$

Step 2. Shaded the region