Question

How do you identify the important parts of `x^2-2x-8=0` to graph it?

Answer 1

x-intercepts: `(4,0)` and `(-2,0)`
vertex: `(1,-9)`

Explanation:

I think the most important parts are the x-intercepts and the vertex.

x-intercepts (The value of `x` when `y=0`)
Given the function `y=x^2-2x-8`, set `y` to `0`. Compute for the x-intercepts by factoring.

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

Now that you've factored it, you can say that the x-intercepts are:

`x-4=0`
`x=4`
`color(blue)((4,0)`

`x+2=0`
`x=-2`
`color(blue)((-2,0)`

Mark the points `(4,0)` and `(-2,0)` on your graph.

vertex
To solve for the vertex, you will have to convert the function into the vertex form `y=a(x-h)^2+k`, where `(h,k)` is the vertex.

`y=x^2-2x-8`
`y+8=x^2-2x`
`y+8+1=x^2-2x+1`
`y+9=(x-1)^2`
`y=(x-1)^2-9`

You can see here that `h=1` and `k=-9`. Therefore, your vertex is `(1,-9)`.

After you have plotted the vertex and x-intercepts, just add a few more points by inserting any value to `x`. This will help to make your graph more accurate.

It should look like this:
graph{x^2-2x-8 [-8.97, 11.03, -9.32, 0.68]}