At how many points will the curves $y=x^2$ and $y=-x^2-2x-1$ intersect in the real $(x,y)$ plane?

Finding Curve Intersection Points

To find the points where the curves $y=x^2$ and $y=-x^2-2x-1$ intersect, we set the expressions for $y$ equal to each other:

$x^2=-x^2-2x-1$

Solving the Quadratic Equation

Rearrange the equation to form a standard quadratic equation ($a{x}^2+bx+c=0$):

$x^2+x^2+2x+1=0$

$2x^2+2x+1=0$

Using the Discriminant

We use the discriminant, $\Delta =b^2-4ac$, to determine the number of real solutions for $x$. A real solution corresponds to an intersection point.

- If $\Delta >0$, there are 2 real solutions (2 intersection points).
- If $\Delta =0$, there is 1 real solution (1 intersection point).
- If $\Delta <0$, there are no real solutions (0 intersection points).

For the equation $2x^2+2x+1=0$, we have $a=2$, $b=2$, and $c=1$. Calculate the discriminant:

$\Delta =(2{)}^2-4(2)(1)$

$\Delta =4-8$

$\Delta =-4$

Conclusion on Intersection Points

Since the discriminant $\Delta =-4$, which is less than 0, there are no real solutions for $x$. Therefore, the two curves do not intersect in the real $(x,y)$ plane.

The number of intersection points is 0.