Question

What is the the vertex of `y = -x^2 - 3`?

Answer 1

`Vertex:(0,-3)`

Explanation:

`y=-x^2-3`
Let us first convert this in vertex from

`color(brown)"vertex form:y=a(x-h)^2+k"`
`color(brown)"vetex:(h,k)"`
Let us write the given equation in vertex form.
`y=(x-0)^2+(-3)`
`Vertex:(0,-3)`

Answer 2

`"vertex"-> (x,y) -> (0,-3)`

Explanation shows what is happening.

Explanation:

Suppose we hade the general equation of `y_1=-x^2`

Then the graph would look like:

Subtract 3 from both sides of the equation. Not only is the equation now `y_1 - 3 = -x^3 - 3` but you have lowered the whole thing by 3.
Let `y_1-3` be written as `y_2` now giving: `y_2=x^2-3`
This graph looks like:

From this you can see that the vertex in the `color(blue)("first case")` is at `x_("vertex")=0" and "y_("vertex")=0` written as `"vertex" ->(x,y) -> (0,0)`

In the `color(blue)("second case")` it has lowered by 3 on the x-axis giving `x_("vertex")=0" and "y_("vertex")=-3` written as

`"vertex"-> (x,y) -> (0,-3)`