# Is this right?

• Jul 15th 2005, 12:48 PM
Hayabusa
Is this right?
Hi,

For y=-2x^2, write the equation of the axis of symmetry and name the vertex.

y=-2x^2

a=(-2) , b= 0 , c= 0

y=(-2)(0) = 0

Then:

x = b/2a

x= 0/2*(-2) = (-0/4)

So y= 0; (0,0)
• Jul 15th 2005, 01:23 PM
ticbol
Half of the answer is wrong.

Vertex is (0,0), allright.

But, axis of symmetry is not y = 0.
Axis of symmetry is x = 0, or the y-axis.

y = -2x^2
It is the x that is squared, so a vertical parabola, so a vertical axis of symmetry, so x=0 or the y-axis.
• Jul 15th 2005, 01:35 PM
Hayabusa
I'm sorry but I do not understand what you are telling me.
• Jul 15th 2005, 01:39 PM
ticbol
Which part you don't understand? All?
• Jul 15th 2005, 05:45 PM
Hayabusa
This is where you lost me.

But, axis of symmetry is not y = 0.
Axis of symmetry is x = 0, or the y-axis.

y = -2x^2
It is the x that is squared, so a vertical parabola, so a vertical axis of symmetry, so x=0 or the y-axis
• Jul 15th 2005, 06:33 PM
ticbol
I see.

So I assume you don't know parabolas yet.

Funny you were told to look for the axis of symmetry of y = -2x^2, which is a vertical parabola that opens down, whose xvertex is at (0,0).

I don't know how to proceed now.

Wait, do you know, or can you graph y = -2x^2?
If you can, do it. And you will see that the graph is symetrical about the y-axis. The y-axis is vertical. The y-axis is also the x=0 vertical line.