Question

How do you find the vertex and intercepts for `12y = x^2 – 6x + 45`?

Answer 1

Vertex is `(3,3)`, `y` intercept is `3 3/4` and there is no `x`-intercept.

Explanation:

In an equation `y=a(x-h)^2+k`, we have vertex as `(h,k)` and intercepts on `y`-axis can be found by putting `x=0` and intercepts on `x`-axis can be found by putting `y=0`. Let us convert the given equation in vertex form.

`12y=x^2-6x+45` can be written as

`y=1/12(x^2-6x)+45/12`

or `y=1/12(x^2-2xx3xx x+3^2-3^2)+45/12`

or `y=1/12(x^2-2xx3xx x+3^2)-9/12+45/12`

or `y=1/12(x-3)^2+36/12`

or `y=1/12(x-3)^2+3`

Hence vertex is `(3,3)`

intercept on `y`-axis is `y=15/4=3 3/4`

Observe that as `a` and `k` are positive, `y` is always greater than `0` and can never be `0`, hence no `x`-intercept.
graph{12y=x^2-6x+45 [-7.12, 12.88, -0.4, 9.6]}