My question: Is it possible to find a specific parabola when you know the average value, and the x intercepts? The parabola I am trying to find has x intercepts at 0 and 18 with an average value of 2.34
Can you reverse engineer the parabola using an average value equation?
In a slightly different notation from @K.defaoite's comment:
If $A>0$ and $a<b$, the parabola $y=A(x-a)(b-x)$ has area from $x=a$ to $x=b$ given by$$A\int_a^b(-ab+(a+b)x-x^2)=A(b-a)^3/6,$$as can be computed easily with $y=\frac{x-a}{b-a}$. The average on this interval is $\mu=A(b-a)^2/6$, so $A=\frac{6\mu}{(b-a)^2}$.