Question

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

Answer 1

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]}