Question

Sketch the following sets of points in the \(x-y\) plane. $$ \left\\{(x, y) \in \mathbb{R}^{2}:\left(y-x^{2}\right)\left(y+x^{2}\right)=0\right\\} $$

Short answer

The solution set for this equation is two parabolas. One is opening upwards at the origin expressed by \(y = x^{2}\) and the other opens downward, also at the origin expressed by \(y = -x^{2}\).

Step-by-step solution

Solve for the first scenario

Begin by focusing on the first condition which is \(y - x^{2} = 0\). This can be rewritten as \(y = x^{2}\). This function produces a parabolic curve, opening upwards, known as a quadratic function. You can plot this curve on the x-y plane.

Solve for the second scenario

The other condition which is \(y + x^{2} = 0\), can be rewritten as \(y = -x^{2}\). Unlike the previous condition, this also yields a parabola but it opens downwards.

Combine the two graphs

Plot both of these functions on the same graph to understand the complete solution set for the given equation. Thus the solution set of the problem will be both the upper and lower halves of the parabola.