Find a function with general shape as a parabola and $f(0) = m$ and $f(1/2) = t$ where $m, t \in \Bbb R$.

Question

I have a relatively simple function design problem that I don't quite remember how to solve. Here are the parameters:

• function of the general shape $-x^2$
• $m$ and $t$ are constants
• $f(0) = m$
• $f(1/2) = t$
• $t$ $>$ $m$

The left side of the parabola should intersect with the y axis @ $m$ and the top / max point should hit at $f(1/2)$ which also $= t$. How would I go about designing this function?

Answer 1

Well, you know that:$$f(x)=t-a(x-1/2)^2\tag{vertex formula}$$and$$f(x)=-ax^2+bx+m\tag{standard form}$$so$$\begin{align}t-a(x-1/2)^2&=-ax^2+ax+t-\frac14a=-ax^2+bx+m\end{align}\\t-\frac14a=m\implies a=4(t-m)\\a=b\implies b=4(t-m)$$

$$f(x)=-4(t-m)x^2+4(t-m)x+m$$