# What are the intercepting points of

y^2=-4x and x^2=y

I set both equations equal to 0, but only get x=0 and there is another intercepting point. please help!

You can solve for one of the equation and substitute it into the other equation. In this case you already have y=x^2 so just plug x^2 into y in the other equation. Once you solve for X ((-4)^(1/3) you can solve for Y (-4)^(2/3)

## To find the intercepting points of the given equations, you correctly started by setting both equations equal to zero. Let's go through the process step by step to ensure we don't miss any solutions.

Equation 1: y^2 = -4x

Equation 2: x^2 = y

First, let's set Equation 1 equal to zero:

y^2 + 4x = 0

Now, let's set Equation 2 equal to zero:

x^2 - y = 0

To find the intercepting points, we need to solve these two equations simultaneously. Let's substitute Equation 2 into Equation 1:

(x^2 - y)^2 + 4x = 0

Expanding the equation:

x^4 - 2x^2y + y^2 + 4x = 0

Since we have y = x^2 (from Equation 2), we can substitute it into the above equation:

x^4 - 2x^2(x^2) + (x^2)^2 + 4x = 0

Simplifying further:

x^4 - 2x^4 + x^4 + 4x = 0

- x^4 + 4x = 0

Factoring out an x:

x(x^3 - 4) = 0

To find the x-values of the intercepting points, we set each factor equal to zero:

x = 0 (One solution)

Next, let's solve x^3 - 4 = 0:

x^3 = 4

x = ∛4 (Cube root of 4)

Now, to find the corresponding y-values, we substitute the x-values into Equation 2:

For x = 0:

y = (0)^2

y = 0

So, we have one intercepting point: (0, 0)

For x = ∛4:

y = (∛4)^2

y = 4^(2/3)

y = 2^(2/3) * 2

Therefore, our second intercepting point is (∛4, 2^(2/3) * 2).

To summarize, the intercepting points of the given equations are:

1. (0, 0)

2. (∛4, 2^(2/3) * 2)