Question

How do you find the intercepts, vertex and graph `f(x)=x^2-9x+9`?

Answer 1

The intercepts are `(0,9)`, `((9+sqrt45)/2,0)` and `((9-sqrt45)/2,0)`
And the vertex is `(9/2,-45/4)`

Explanation:

Let's rewrite `f(x)` in the vertex form
So, `f(x)=x^2-9x+9=x^2-9x+81/4+9-81/4`
`f(x)=(x-9/2)^2-45/4`

So the vertex is `(9/2,-45/4)`
The y intercepts are `x=0` `=>` `y=9`

The x intercepts are when `y=0`
`f(x)=x^2-9x+9=0`
so we calculate `Delta=b^2-4ac=81-4*9*1=45`
and we get `x=(9+-sqrt45)/2`
So the x intercepsts are `((9+sqrt45)/2,0)` and `((9-sqrt45)/2,0)`
You can also use `f(x)=0`
`(x-9/2)^2-45/4=0`
`(x-9/2)^2=45/4`
`(x-9/2)=+-sqrt45/2`
`x=9/2+-sqrt45/2=(9+-sqrt45)/2`

graph{x^2-9x+9 [-28.86, 28.85, -14.43, 14.43]}