Question

How do you solve `x^2 - 18x + 81 = 0` graphically?

Answer 1

The vertex is `(9,0)`, and there are no stretch factors.

Explanation:

We begin with `x^2-18x+81=0`. For me, I immediately think of "how can I gat this out of standard form and into vertex or intercept form?"

Well, for vertex form, we could "complete the square". I'm a big fan of this method, but I always got to factoring first, and if that doesn't work I move on to other methods like completing the square or the quadratic formula.

So, let's begin with factoring.

We are trying to factor `x^2-18x+81`, and we are looking for something that multiplys to `81`, and adds to `-18`.

So, the factors of `81` are:
`+-1` and `+-81`
`+-3` and `+-27`
`+-9` and `+-9`

No, out of these factors we are supposed to have them add to `-18`
So,
`+-1` + `+-81` = `80, 82, -80, -82`
`+-3` + `+-27` = `30, -30, 24, -24`
`+-9` + `+-9` = `18, 0, 0, color(green)(-18)`

There it is. `-9` and `-9` add to `-18` and multiply to `81`!

So, that means we can rewrite `x^2-18x+81` as `(x-9)(x-9)`, or `(x-9)^2`. If we wanted to write this in vertex form, we just write this: `(x-9)^2+0`.

No, look: there is no stretch factor for `(x-9)^2+0`, and we know that the vertex is `9,0`. Just graph that vertex, and have it open upwards, with no stretching.

To find other points, just replace `x` with the number your trying to find, like this: `((8)-9)^2+0`, which is `(-1)^2+0`, or `1`. So, we know another point: `(8,1)`.

If we solve for when `x` is `-1`, we have `((10)-9)^2+0` or `(1)^2+0`, which is just `1`, or `(10,1)`.

Now you have the vertex and a couple of point to help guide you as you draw your graph

If you want to double check this, just graph it

graph{y=(x-9)^2+0}