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.