Question

How do you find the vertex and intercepts for `y=x^2 – 12x + 36`?

Replaced equation (with 2 solutions for `x`) into a corresponding function with vertex and intercepts.

Answer 1

Vertex at `(6,0)`
y-intercept: `36`
x-intercept: 6#

Explanation:

`y=x^2-12x+36`

`rArr y = (x-6)^2+0`
this is the vertex form of a parabola with vertex at `(6,0)`

The y-intercept is the value of `y` when `x=0`
`color(white)("XXX")y=(0)^2-12(0)+36 = 36`

The x-intercept is the value of `x` when `y=0`
`color(white)("XXX")0=x^2-12x+36 = (x-6)(x-6)`
`color(white)("XXX")rArr x=6` (single intercept point)
graph{x^2-12x+36 [-25.27, 54.73, -1.6, 38.4]}