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

Short 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

Step-by-step solution

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

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

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

$x^2+2x=0.$

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

$x=0orx+2=0.$

$x=-2.$

When$x=0orx=-2$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