Answer
See explanations.
Work Step by Step
Step 1. Since $(0,y_1)$ is on $y=ax^2+bx+c$, we have $y_1=c$
Step 2. Similarly for $(-h,y_0)$ and $(h,y_2)$, we have $y_0=a(-h)^2+b(-h)+c$ and $y_2=ah^2+bh+c$
Step 3. Add up the two equations in Step 2 to get $y_0+y_2=2ah^2+2c$
Step 4. With the given area formula $Area=\frac{h}{3}(2ah^2+6c)$, we have $Area=\frac{h}{3}(2ah^2+2c+4c)=\frac{h}{3}(y_0+y_2+4y_1)$
