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

Mar 31, 2017

#### Explanation:

This is a typical quadratic equation, which in vertex form appears as $y = a {\left(x - h\right)}^{2} + k$, which is a parabola with axis of symmetry as $x = h$ and vertex at $\left(h , k\right)$, 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 $y + {x}^{2} = 8 x - 12$, let us convert it into vertex form. The equation becomes

$y = - {x}^{2} + 8 x - 12$

i.e. $y = - 1 \left({x}^{2} - 2 \times 4 \times x + {4}^{2} - {4}^{2}\right) - 12$

or $y = - 1 {\left(x - 4\right)}^{2} + 16 - 12$

or $y = - 1 {\left(x - 4\right)}^{2} + 4$

Now as $a = - 1$, we have a maxima at $\left(4 , 4\right)$

axis of symmetry is $x = 4$

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

Corresponding value of $y$ will be $y = \left\{- 60 , - 32 , - 12 , 0 , 0 , - 12 , - 32 , - 60\right\}$

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