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 whether the resulting quadratic has real roots.

Answer 1

The Correct Option is A

Step 1: Form the equation for meeting points.
The two curves meet where their $y$ values agree, so
\[ x^2=-x^2-2x-1 \]
which rearranges to
\[ 2x^2+2x+1=0 \]
Step 2: Divide through to simplify.
Divide every term by $2$:
\[ x^2+x+\frac{1}{2}=0 \]
Step 3: Complete the square.
\[ x^2+x=\left(x+\frac{1}{2}\right)^2-\frac{1}{4} \]
So the equation becomes
\[ \left(x+\frac{1}{2}\right)^2-\frac{1}{4}+\frac{1}{2}=0 \]
\[ \left(x+\frac{1}{2}\right)^2=-\frac{1}{4} \]
Step 4: Check if a real solution exists.
A square of a real number can never be negative, but the right side here is $-\frac{1}{4}$, which is negative. So no real value of $x$ can satisfy this equation.
Step 5: Conclude the number of intersection points.
Since there is no real $x$ where the curves agree, they do not cross anywhere in the real plane.
\[ \boxed{0} \]