# What is the axis of symmetry and vertex for the graph y = x^2 – 16x + 58?

Sep 4, 2017

The vertex form of a quadratic equation like this is written:

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

...if we can rewrite the initial equation in this form, the vertex coordinates can be read directly as (h,k).

Converting the initial equation to vertex form requires the infamous "completing the square" maneuver.

If you do enough of these, you begin to spot patterns. For example, -16 is $2 \cdot - 8$, and $- {8}^{2} = 64$. So if you could convert this to an equation that looked like ${x}^{2} - 16 x + 64$, you'd have a perfect square.

We can do this via the trick of adding 6 and subtracting 6 from the original equation.

$y = {x}^{2} - 16 x + 58 + 6 - 6$

$= {x}^{2} - 16 x + 64 - 6$

$= {\left(x - 8\right)}^{2} - 6$

...and bam. We have the equation in vertex form. a = 1, h = 8, k = -6 Vertex coordinates are (8, -6)

The axis of symmetry is given by the x coordinate of the vertex. I.e., the axis of symmetry is the vertical line at x = 8.

It's always handy to have a graph of the function as a "sanity check".

graph{x^2 - 16x + 58 [-3.79, 16.21, -8, 2]}

GOOD LUCK!