Thread: Need help finding intercepts

Hi there. I am trying to test an equation for symmetry. The equation is x^2 = y^3 + 4. Before I can test it for symmetry, I need to find the intercepts. This is where I'm having trouble.

I think I found the intercepts for x:
(2,0), (-2,0)

I am having trouble finding the intercepts for y.

I substitute 0 for X and get 0 + y^3 + 4. I convert that to y^3 = -4. I am not sure where to go from there.

Can I get a cube root of negative four? Solving for y, would my answer be the cube root of four i? Does that mean there is no y intercept?

Thanks for any help, I'm not sure why I'm struggling so much with this. I think I forgot some basic stuff during summer vacation.

Originally Posted by juliagregs

Hi there. I am trying to test an equation for symmetry. The equation is x^2 = y^3 + 4. Before I can test it for symmetry, I need to find the intercepts. This is where I'm having trouble.

I think I found the intercepts for x:
(2,0), (-2,0)

I am having trouble finding the intercepts for y.

I substitute 0 for X and get 0 + y^3 + 4. I convert that to y^3 = -4. I am not sure where to go from there.

Can I get a cube root of negative four? Solving for y, would my answer be the cube root of four i? Does that mean there is no y intercept?

Thanks for any help, I'm not sure why I'm struggling so much with this. I think I forgot some basic stuff during summer vacation.

Yes you can take the cube root of negative numbers and get a real answer. When you cube a number, you are multiplying this number by itself 3 times. This give negative x negative x negative, which gives negative. From this we get .

Thank you!

Here is the property of the parabola
y=-x^(2)+4x-3
a = -1, b =4, c = - 3

-b/(2a) = - 4/[2(-1)] = 2

f(2) = - (2)^2 + 4(2) - 3 = 1

Use the standard form to determine the vertex and x-y intercepts
1to1Tutor