Looking for a familiy of quadratic functions with vertices along a line

Question

I am looking for a familiy of quadratic functions with the following characteristics:

• the vertex is at y=1 and in the interval [2,4] for x
• the parabola runs through (0,0)

So my "leftmost" function is $-\frac{x^2}{4} +x $ and the "rightmost" one is $-\frac{x^2}{16} +\frac{x}{2} $.

Now I have a parameter z, which takes values between 0 and 99 (or 100 if that is more convenient). For z=0 I want the rightmost function, for z=99 the leftmost one. With growing z the vertices should move along the line y=1. Equal spacing would be fine, also closer spacing for low z and wider for bigger z.

The naive approach of $-\frac{x^2}{16-12\cdot\frac{z}{99}} +\frac{x}{2-\frac{z}{99}} $ does not work, because the vertices leave the y=1 line.

Answer 1

I think this works . . .

For $0 \le z \le 99$, let $f_z(x)$ be defined by $$f_z(x) = a(z)x^2 + b(z)x$$ where $a,b$ are given by \begin{align*} a(z)&=-\frac{9801}{4(198-z)^2}\\[4pt] b(z)&=\frac{99(792-4z)}{4(198-z)^2}\\[4pt] \end{align*} Then identically, \begin{align*} f_0(x)&=-\frac{1}{16}x^2+\frac{1}{2}x\\[4pt] f_{99}(x)&=-\frac{1}{4}x^2+x\\[4pt] \end{align*} and for each $z \in [0,99]$, the graph of $f$

• passes through $(0,0)$.$\\[2pt]$
• has vertex at $(h(z),1)$, where $h(z) = 4 - {\large{\frac{2z}{99}}}$.