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 both equations equal, form a quadratic in x, and check its discriminant.

Answer 1

The Correct Option is A

Step 1: Look at each curve as a parabola.
The curve $y = x^2$ opens upward with its lowest point at the origin $(0,0)$.
The curve $y = -x^2 - 2x - 1$ opens downward, since the coefficient of $x^2$ is negative.

Step 2: Find the vertex of the second parabola.
Rewrite $y = -x^2 - 2x - 1$ by factoring out the negative sign:
\[ y = -(x^2 + 2x + 1) = -(x+1)^2 \]
This is a perfect square, so the vertex sits at $x = -1$, $y = 0$, and this is the highest point the curve ever reaches.

Step 3: Compare the highest and lowest values.
For the downward parabola $y = -(x+1)^2$, the value of $y$ is always less than or equal to $0$, since a square is never negative.
For the upward parabola $y = x^2$, the value of $y$ is always greater than or equal to $0$, so the two curves can only share a point where both equal $0$ at the same time.

Step 4: Check where each curve equals zero.
$y = x^2 = 0$ only at $x = 0$.
$y = -(x+1)^2 = 0$ only at $x = -1$.
These two zero points happen at different $x$ values, so the curves never touch $y = 0$ together.

Final Answer:
Since the curves cannot meet even at their shared boundary value $y = 0$, they never cross in the real plane. \[ \boxed{0} \]