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

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}