# Thread: intersection points of 2 curves

1. ## intersection points of 2 curves

Hi everybody, this may be really simple and im just missing the point. But how can you calculate the 2 intersection points of 2 quadratics if one is x^2 and the other is -x^2?

Many thanks

Daniel

2. Originally Posted by wilsnunn
Hi everybody, this may be really simple and im just missing the point. But how can you calculate the 2 intersection points of 2 quadratics if one is x^2 and the other is -x^2?

Many thanks

Daniel
Hi Daniel,

The graph of $f(x)=x^2$ is a parabola with vertex at the origin and opening upward.

The graph of $f(x)=-x^2$ is a parabola with vertex at the origin and opening downward.

Their obvious intersection is the origin.

3. Originally Posted by wilsnunn
Hi everybody, this may be really simple and im just missing the point. But how can you calculate the 2 intersection points of 2 quadratics if one is x^2 and the other is -x^2?

Many thanks

Daniel
There is only one intersection point - at the origin.

You can find this by setting them equal to each other

$-x^2 = x^2$

$2x^2 = 0$

$x = 0$

4. Originally Posted by masters
Hi Daniel,

The graph of $f(x)=x^2$ is a parabola with vertex at the origin and opening upward.

The graph of $f(x)=-x^2$ is a parabola with vertex at the origin and opening downward.

Their obvious intersection is the origin.
I meant if one was say $f(x)=x^2-6x+9$ and the other was $f(x)=-x^2+7x-11$

How would you do it then? Sorry I should have mentioned that in the opening thread.

Daniel

5. Originally Posted by wilsnunn
I meant if one was say $f(x)=x^2-6x+9$ and the other was $f(x)=-x^2+7x-11$

How would you do it then? Sorry I should have mentioned that in the opening thread.

Daniel
$f(x)$ is just another name for y

let $y = y$

=> $x^{2} - 6x + 9 = -x^{2} +7x - 11$
=> $2x^{2} - 13x + 20 = 0$

solve that quadratic for x and find the corresponding y value by subbing x into original equation of curve