# How do you find the vertex of y= x^2 - 16x + 73?

Jun 5, 2018

$\left(8 , 9\right)$

#### Explanation:

Use the equation of a vertex:

If you have a quadratic equation $y = A {x}^{2} + B x + C$, then the x coordinate of the vertex is found by using the formula:

$- \frac{B}{2 A}$

For your equation, $A = 1 , B = - 16 , C = 73$

So $- \frac{B}{2 A} = - \frac{- 16}{2 \cdot 1} = 8$

Thus we know the vertex has an x-coordinate of 8.

To find the y-coordinate, just plug in the x-coordinate back into the function:

$y = {\left(8\right)}^{2} - 16 \left(8\right) + 73$
$y = 64 - 128 + 73$
$y = 9$

So our vertex has a y-coordinate of 9. Therefore our vertex is $\left(8 , 9\right)$

An alternate way to approach this is to first put the equation into vertex form. See below for more:

#### Explanation:

The general form for the vertex form is:

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

And we get there by completing the square:

$y = {x}^{2} - 16 x + 73$

We take the $- 16$, halve it, square the half, then add and subtract it:

$y = {x}^{2} - 16 x + 64 - 64 + 73$

Notice that we can take the first 3 terms and write it as a quantity squared:

$y = {\left(x - 8\right)}^{2} - 64 + 73$

and we can simplify the other two terms:

$y = {\left(x - 8\right)}^{2} + 9$

With this form, the vertex is given as $\left(h , k\right)$. Here it's $\left(8 , 9\right)$ (remember that in the general form, we have $- h$ and so $h$ values are positive numbers and vice versa.