Graphing Parabolas ( finding area)

Question

Find the area of the triangle formed by the origin and the points of intersection of parabolas $y=−3x^2+20$ and $y=x^2−16$.

I tried graphing it. I could figure out what it meant when it's said formed by the origin. Please help.

Could someone also please post a picture of the two graphs using online because I don't know how to do it.

Answer 1

Find solutions to $$\begin{cases} y=-3x^2+20\\ y=x^2-16 \end{cases}$$

$$-3x^2+20=x^2-16$$ $$36=4x^2$$ $$9=x^2$$ $$x=\pm 3\implies y=9-16=-7$$

So the solutions are $(3,-7)$ and $(-3,-7)$. The area is equal to

$$S=\frac{6\cdot 7}{2}=21$$

Answer 2

Just to illustrate the answer of kamilo9875.

In red the parabola $ y=20-3x^2$ in blue $y=x^2-16$, The triangle is $AOB$ and $A=(-3,-7)$, so the area is $3\times 7=21$.