http://www.webassign.net/cgi-bin/sym...29%2A%2A2%20-3

How do I find the vertex?

I set y to zero then I get -3? I think.

Printable View

- Jul 16th 2011, 12:08 PMtheloserInverted Parabola
http://www.webassign.net/cgi-bin/sym...29%2A%2A2%20-3

How do I find the vertex?

I set y to zero then I get -3? I think. - Jul 16th 2011, 12:35 PMSpringFan25Re: Inverted Parabola
Can you find the vertex of $\displaystyle x=y^2$ ?

Your function is stretched and shifted version of $\displaystyle x=y^2$. Identify the transformation and you will see where the new vertex is. - Jul 16th 2011, 04:41 PMHallsofIvyRe: 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, $\displaystyle x= 3(y+1)^2- 3$, is never less than -3. It**is**-3 when the**square**is 0- that is, when y+ 1= 0. - Jul 16th 2011, 06:01 PMskokerRe: Inverted Parabola
it helps to write out the standard form formula. then you can see what its telling you.

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

you also know your values are inverted.