Question

What is the the vertex of `y =x^2-6x-7`?

Answer 1

`P(3,-16)`

Explanation:

There are different ways this can be done.

This equation is in standard form, so you can use the formula `P(h,k) = (-b/(2a),-d/(4a))` Where the (d) is the discriminant. `d = b^2-4ac`

Or to save time, you can find the (x) coordinat for the vertex with `-b/(2a)` and put the result back in to find the (y) coordinat.

Alternatively, you can rearrenge the equation into vertex form:
`a(x-h)^2+k`

To do this start by putting a outside the brackets. This is easy because `a=1`

`x^2-6x-7 = 1(x^2-6x) - 7`

Now we have to change `x^2-6x` into `(x-h)^2`
To do this we can use the quadratic sentence: `(q-p)^2 = q^2+p^2-2qp`

Let's say `q=x` therefore we get:
`(x-p)^2 = x^2+p^2-2xp`

This looks sort of what we need, but we are still far of, as we only have `x^2`.

If we look at `x^2-6x`, we can se that there is only one part raised to the power of two, therefore `p^2` must be removed. This means:

`(x-p)^2-p^2=x^2-2xp`

Looking at the right side, we can see it is almost `x^2-6x`, in fact we only have to solve `-2xp = -6x` `iff p = 3`

This means:
`(x-3)^2-9 = x^2-6x`

Another way of doing it would be to make a qualified guess and use the quadratic sentences to see if it is correct.

Now go back to our original formula and replace `x^2-6x` with `(x-3)^2-9`

We get:

`1(x^2-6x) - 7 = 1((x - 3)^2-9) - 7 = 1(x - 3)^2-9 - 7 = 1(x - 3)^2-16`
This is similar to the vertex form:
`a(x-h)^2+k`
Where
`h = 3` and `k=-16`

When the quadratic equation is in vertex form, the vertex is simply the point `P(h,k)`

Therefore the vertex is `P(3,-16)`