1. Inverted Parabola

How do I find the vertex?
I set y to zero then I get -3? I think.

2. Re: Inverted Parabola

Can you find the vertex of $x=y^2$ ?

Your function is stretched and shifted version of $x=y^2$. Identify the transformation and you will see where the new vertex is.

3. Re: Inverted Parabola

No, you do NOT set y equal to 0. That would give the y-intercept which, in general, has nothing to with the vertex.

Since a square is never 0, $x= 3(y+1)^2- 3$, is never less than -3. It is -3 when the square is 0- that is, when y+ 1= 0.

4. Re: Inverted Parabola

it helps to write out the standard form formula. then you can see what its telling you.

$f(x)=a(x-h)^2+k$ vertex=(h,k)

you also know your values are inverted.