Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Set the two expressions for y equal and check the discriminant of the resulting quadratic.

Answer 1

The Correct Option is A

Step 1: Rewrite the second curve to see its shape.
$y = x^2$ is an upward-opening parabola with its lowest point (vertex) at the origin $(0,0)$, so every point on it has $y \geq 0$.
The second curve is $y = -x^2 - 2x - 1 = -(x^2+2x+1) = -(x+1)^2$. This is a downward-opening parabola, and its highest possible value of $y$ (the vertex) occurs at $x=-1$, giving $y = -(0)^2 = 0$. So every point on it has $y \leq 0$.

Step 2: Compare the two curves using their extreme values.
For $y=x^2$, $y \geq 0$ always, with equality only at $x=0$.
For $y=-(x+1)^2$, $y \leq 0$ always, with equality only at $x=-1$.

Step 3: Find where both conditions can hold together.
For an intersection point, both $y \geq 0$ and $y \leq 0$ must hold at the same time, which forces $y = 0$. From the first curve, $y=0$ only at $x=0$. From the second curve, $y=0$ only at $x=-1$. Since these two $x$-values are different, no single point sits on both curves at $y=0$, and no other $y$ value can satisfy both inequalities together.

Step 4: Conclude.
The two parabolas open in opposite directions, and their vertices, the only points where each one touches $y=0$, sit at different $x$-values ($x=0$ and $x=-1$). So the curves never cross. \[ \boxed{0} \]