At how many points will the curves y = x² and y = −x² −2x−1 intersect in the real (x,y) plane?

The correct option is 0.

To find the number of points at which the curves intersect in the real (x, y) plane, we set the equations of the two curves equal to each other:

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

Next, we move all the terms to one side of the equation to form a standard quadratic equation:

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

Combining the like terms gives:

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

This is a quadratic equation of the form $a{x}^2+bx+c=0$, where:
$a=2$,
$b=2$, and
$c=1$.

To determine the number of real roots (which correspond to the number of real intersection points), we calculate the discriminant, $D$:

$$D=b^2-4ac$$

Substituting the values of $a$, $b$, and $c$ into the formula:

$$D=2^2-4\left(2\right)\left(1\right)$$

$$D=4-8$$

$$D=-4$$

Since the discriminant is negative ($D<0$), the quadratic equation has no real solutions. Therefore, the curves do not intersect at any point in the real plane.