What is the vertex of `y=5(x+3)^2-9`?

There are two ways to solve it:

1) Quadratics:

For the equation `ax^2+bx+c=y` :

The `x`-value of the vertex `=(-b)/(2a)`

The `y`-value can be found out by solving the equation.

So now, we have to expand the equation we have to get it in quadratic form:

`5(x+3)^2-9=y`

`-> 5(x+3)(x+3)-9=y`

`-> 5(x^2+6x+9)-9=y`

`-> 5x^2+30x+45-9=y`

`-> 5x^2+30x+36=y`

Now, `a=5` and `b=30`. (FYI, `c=36`)

`-> (-b)/(2a)=(-(30))/(2(5))`

`->(-b)/(2a) = (-30)/10`

`->(-b)/(2a) = -3`

Thus, the `x`-value `=-3`. Now, we substitute `-3` for `x` to get the `y` value of the vertex:

`5x^2+30x+36=y`

becomes:

`5(-3)^2+30(-3)+36=y`

`-> 45+(-90)+36=y`

`-> y=81-90`

`-> y=-9`

Thus, since `x=-3` and `y=-9`, the vertex is:

`(-3 , -9)`

2) This is the easier way of doing it - by using the Vertex Formula:

In the equation `a(x-h)^2+k=y`, the vertex is `(h,k)`

We are already given an equation in the Vertex format, so it is easy to find out the Vertex coordinates:

`5(x+3)^2-9=y`

can be rewritten as:

`5(x-(-3))^2-9=y`

Now we have it in the Vertex-form, where `h=-3`, and `k=-9`

So, the Vertex coordinates are:

`(h,k)`

`=(-3,-9)`

Tip: you can change an equation in a quadratic form to a vertex form by completing the square. If you are not aware of this concept, search it up on the Internet or post a question on Socratic.