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

2 Answers
Jun 5, 2018

(8,9)

Explanation:

Use the equation of a vertex:

If you have a quadratic equation y=Ax^2+Bx+C, then the x coordinate of the vertex is found by using the formula:

-B/(2A)

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

So -B/(2A)=-(-16)/(2*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=(8)^2-16(8)+73
y=64-128+73
y=9

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

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=(x-h)^2+k

And we get there by completing the square:

y=x^2-16x+73

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

y=x^2-16x+64-64+73

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

y=(x-8)^2-64+73

and we can simplify the other two terms:

y=(x-8)^2+9

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