Question

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

Answer 1

`(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)`

Answer 2

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.