How do you solve x^2-2x-8=0 by graphing?

1 Answer
Jun 24, 2018

We need to find the vertex, x-intercepts, and y-intercept to give us an idea of what the graph will look like.

Explanation:

Let's find the vertex first. We can find the x-coordinate by using x=-b/(2a):

x=-2/(2*1)=-2/2=1

The y-coordinate can be found by plugging the x-value back into the equation:

(1)^2-2(1)-8 = 1-2-8 = 9

The vertex is (1,9). Next, let's find the x-intercepts, which are the zeroes of the equation. Let's factor the polynomial and find them:

x^2-2x-8=0
(x-4)(x+2)=0
x-4=0
x=4
x+2=0
x=-2

The x-intercepts are (4,0) and (-2,0). Finally, the y-intercept can be found by letting x=0:

0^2-2(0)-8=-8

The y-intercept is (0,-8). You can plot the bolded points, which will be able to create a decent outline of the parabola, If you need them, you can plot other points on the graph, such as (2,-8), (-1,-5), and (3,-5). This is what the graph will look like:

graph{x^2-2x-8 [-9, 11, -10, 1]}