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]}