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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.

Peter Leupold
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1 Answers1

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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 $h(z) = 4 - {\large{\frac{2z}{99}}}$.
quasi
  • 58,772
  • Thank you, this seems to work indeed, and the vertices are spaced out equally. I figured that the z had to enter in a "quadratic manner" to influence the quadratic x in a steady manner. Have you used the formula for the discriminant to deduce this? – Peter Leupold Dec 01 '17 at 14:01
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    No, I just used the standard form $f(x) = a(x-h)^2 + k$, with $k=1$. I forced $h$ to be equally spaced, and solved for $a$ based on the requirement that the graphs pass through the origin. – quasi Dec 01 '17 at 14:04