Q. 84 - Precalculus Enhanced with Graphing Utilities

Question

Suppose that $f\left(x\right)=x^2+2x-8,$

(a) What is the vertex of $f$ ?

 (b) What are the $x$-intercepts of the graph of $f$ ?

(c) Solve $f\left(x\right)=-8$for $x,$. What points are on the graph of $f$?

(d) Use the information obtained in parts (a) – (c) to graph $f\left(x\right)=x^2+2x-8.$

Step-by-Step Solution

Answer

(a) Vertex is $\left(-1,-9\right).$

(b) $x$- intercept is $x=2,-9.$

(c) The point  is on the graph of $\left(0,-8\right),\left(-2,8\right).$

(d) The graph of 

Part (a) Step 1. Given information

It is given that $f\left(x\right)=x^2+2x-8,$ we need to determine the vertex of $f.$

Step 2. Simplify

$f\left(x\right)=x^2+2x-8.$

$f\left(x\right)=x^2+2x+1-1-8.$

       $={\left(x+1\right)}^2-9.$

The vertex is $\left(-1,-9\right).$

Part (b) Step 1. Given information

It is given that $f\left(x\right)=x^2+2x-8,$ we need to determine the $x$- intercept of the graph.

Step 2. Simplify

Put $f\left(x\right)=0.$

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

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

   $=\frac{-2\pm \sqrt{4+32}}{2}.$

    $=\frac{-2\pm \sqrt{36}}{2}.$

    $=\frac{-2\pm 6}{2}.$

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

The $x$- intercept is $x=2,-4.$

Part (c) Step 1. Given information

It is given that $f\left(x\right)=x^2+2x-8,$ we need to solve $f\left(x\right)=-8$ for $x$ and also the points on the graph.

Step 2. Simplify

$f\left(x\right)=-8.$

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

$x^2+2x=0.$

$x\left(x+2\right)=0.$

$x=0 or x+2=0.$

                  $x=-2.$

When $x=0 or x=-2$ the value of the function is $-8,$ the points $\left(0,-8\right),\left(-2,-8\right)$ are on the graph of $f.$

Part (d) Step 1. Given information

It is given that $f\left(x\right)=x^2+2x-8,$ we need to use the information of the part a-c to graph $f.$

Step 2. Graph of f

The graph of