Question

How do you identify the important parts of `y + x^2 = 8x − 12` to graph it?

Answer 1

Please see below.

Explanation:

This is a typical quadratic equation, which in vertex form appears as `y=a(x-h)^2+k`, which is a parabola with axis of symmetry as `x=h` and vertex at `(h,k)`, where we have a maxima or minima depending on whether `a` is negative or positive.

In such cases, we need to identify

1. first the axis of symmetry

2. then the vertex and finally

3. a few points on both sides of axis of symmetry.

Hence, let us first convert it into vertex form. As we have `y+x^2=8x-12`, let us convert it into vertex form. The equation becomes

`y=-x^2+8x-12`

i.e. `y=-1(x^2-2xx4xx x+4^2-4^2)-12`

or `y=-1(x-4)^2+16-12`

or `y=-1(x-4)^2+4`

Now as `a=-1`, we have a maxima at `(4,4)`

axis of symmetry is `x=4`

Let us choose points with `x={-4,-2,0,2,6,8,10,12}`
(around `x=4`)

Corresponding value of `y` will be `y={-60,-32,-12,0,0,-12,-32,-60}`

and graph appears as follows:
graph{y+x^2=8x-12 [-6, 14, -70, 10]}