Question

Help me with this question.

Screenshot attached.

Answer 1

open your eyes.

Answer 2

I know you are.

Answer 3

To clarify, the leftmost figure doesn't apply here, right? Just the one on the right?

Answer 4

Assuming that's the case...

You're given a vertex of \(\left(\dfrac{b}{2},h\right)\), which means the equation of the parabola, in vertex form, is
\[y=a\left(x-\frac{b}{2}\right)^2+h\]
To solve for \(a\), you have to use the given points \((b,0)\). You have
\[0=a\left(b-\frac{b}{2}\right)^2+h\\
0=\frac{ab^2}{4}+h\\
a=-\frac{4h}{b^2}\]
So your parabola's equation is
\[y=-\frac{4h}{b^2}\left(x-\frac{b}{2}\right)^2+h\]
I'm not sure what class this is for, so I don't know what method you have to use for finding the area.