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 ($ax^2 + bx + c = 0$):

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

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

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.