Question

How do you find the vertex of `y=-x^2+4x+12`?

Answer 1

Factor the equation such that your get `y=-(x-2)^2+16` and your vertex appears at (2,16)

Explanation:

when you factor a quadratic to find the vertex, it takes the form of:

`y=a(x-h)^2+k`, where your vertex is `(h,k)`

Note that `a` is the same `a` as the original quadratic equation. For this equation:

`y=(-1)(x-h)^2+k rArr y=(-1)(x^2-2xh+h^2)+k`

`y=-x^2+2xh-h^2+k`

If we rearrange the above equation to satisfy the other two terms in the original quadratic:

`4x=2xh`
and
`12=k-h^2`

Solving the first equation:
`4=2h rArr color(red)(h=2)`

...and now the second:
`12=k-(2)^2 rArr 12=k-4 rArr color(blue)(k=16)`

Now we have our factors for the vertex form of the quadratic, and we have our vertex!

`y=-(x-2)^2+16`
`"Vertex"=(2,16)`

There are other, quicker ways to compute this too:

`h=(-b)/(2a)`, then plug `h` into the quadratic to find `k`

If you use calculus, you can take the derivative of the quadratic to find `h`, the vertex is where the slope is 0 so we can find h by plugging in where the derivative goes to 0... although this is literally the same as using `h=(-b)/(2a)`.

`(dy)/(dx)=2ax+b`

`0=2ah+b rArr h=(-b)/(2a)`