Question

What is the vertex of `y= -(x-6)^2-4x^2-2x-2`?

Answer 1

`(1,-33)`

Explanation:

We start out with `y=-(x-6)^2-4x^2-2x-2`.

The first thing we want to do is combine like terms, but there aren't any... yet . We need to expand `(x-6)^2`, which we do by rewriting it as `(x-6)*(x-6)` and multiply through to create `x^2-12x+36`.

We plug that into where `(x-6)^2` used to be, and we see this: `y=-(x^2-12x+36)-4x^2-2x-2`. Distribute the `-` into the `(x^2-12x+36)`, changing it to `-x^2+12x-36-4x^2-2x-2`.

NOW we can combine like terms.

`-x^2-4x^2` becomes `-5x^2`
`12x-2x` becomes `10x`
`-36-2` becomes `-38`.
Put it all together and we have `-5x^2+10x-38`. This is not factorable, so we will solve by completing the square. To do that, the coefficient of `x^2` must be 1, so we factor out `-5`. The equation now becomes `-5(x^2-2x+38/5)`. To complete the square, we have to find the value that will make `x^2-2x` factorable. We do that by taking the middle term, `-2x`, dividing it by two (`-2/2 = -1`), and squaring the answer you got (`-1^2=1`).

We then rewrite the equation as `y=-5(x^2-2x+1+38/5)`.

But wait!

We can't just stick a random number in the equation! What we do to one side we must do to the other. Now, I don't know about you, but I don't really want to change `y`. I like having it isolated, but we still have to deal with adding a `1` to only one side of the equation.

But you know, we could just subtract an `-1` , which would cancel out the `1` so it wouldn't effect the equation. Let's do that!

Now the equation reads: `y=-5(x^2-2xcolor(red)(+1-1)+38/5)`. We can simplify `x^2-2x+1` to `(x-1)^2` and simplify `-1+35/5` to just `33/5`. We can simplify the equation to `-5((x-1)^2+33/5)`. The last step is to multiply the `-5 * 33/5`, and because the `5`s divide out (like so: `-cancel(5)*(33/cancel(5))`), all that is left is -33.

Putting it all together, we have `y=-5(x-1)^2-33`.

This is actually in vertex form. All we have to do to find the vertex is take the `y=-5(xcolor(red)(-1))^2color(blue)(-33)` and put it into coordinate-pair form: `(color(red)(1),color(blue)(-33))`.

NOTE the `color(red)(x)` value changed signs once I took it out of the equation. Remember this as it happens every time.