# How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. y=x^2-16x + 63?

May 7, 2015

$y = {x}^{2} - 16 x + 63$

1. Has a y-intercept at $y = 63$ as determined by setting $x = 0$ (and, yes; I know this wasn't asked for but it might help in graphing)

2. Since $\left({x}^{2} - 16 x + 63\right)$ can be factored as
$\left(x - 7\right) \left(x - 9\right)$, the x-intercepts are at $7$ and $9$

3. The vertex occurs at the point where $y ' = 0$
$y ' = 2 x - 16 = 0 \rightarrow x = 8$ (This could also be determined as the mid x coordinate between the two x-intercepts).
At $x = 8$
$y = {8}^{2} - 16 \left(8\right) + 63 = 0$
So the vertex is at $\left(8 , 0\right)$

4. This equation is that of a standard parabola so the axis of symmetry is the vertical line through the vertex:
$x = 8$
graph{x^2-16x+63 [-5.47, 22.99, -2, 12.25]}

I'm not very good at drawing smooth curves so I've used an external tool; the labeling is up to you)