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

Mar 16, 2016

The vertex is $\left(9 , 0\right)$, and there are no stretch factors.

#### Explanation:

We begin with ${x}^{2} - 18 x + 81 = 0$. 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 ${x}^{2} - 18 x + 81$, and we are looking for something that multiplys to $81$, and adds to $- 18$.

So, the factors of $81$ are:
$\pm 1$ and $\pm 81$
$\pm 3$ and $\pm 27$
$\pm 9$ and $\pm 9$

No, out of these factors we are supposed to have them add to $- 18$
So,
$\pm 1$ + $\pm 81$ = $80 , 82 , - 80 , - 82$
$\pm 3$ + $\pm 27$ = $30 , - 30 , 24 , - 24$
$\pm 9$ + $\pm 9$ = $18 , 0 , 0 , \textcolor{g r e e n}{- 18}$

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

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

No, look: there is no stretch factor for ${\left(x - 9\right)}^{2} + 0$, 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: ${\left(\left(8\right) - 9\right)}^{2} + 0$, which is ${\left(- 1\right)}^{2} + 0$, or $1$. So, we know another point: $\left(8 , 1\right)$.

If we solve for when $x$ is $- 1$, we have ${\left(\left(10\right) - 9\right)}^{2} + 0$ or ${\left(1\right)}^{2} + 0$, which is just $1$, or $\left(10 , 1\right)$.

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}