# Problem of the Week

Problem E and Solution

Reach for the Sky

## Problem

The equation \(y=-5x^2+ax+b\), where \(a\) and \(b\) are real numbers and \(a \ne b\), represents a parabola. If this parabola passes through the points with coordinates \((a,b)\) and \((b,a)\), determine the maximum value of the parabola.

## Solution

Since \((a,b)\) lies on the parabola, it satisfies the equation of the parabola. We can substitute \(x=a\) and \(y=b\) into the equation \(y=-5x^2+ax+b\). \[\begin{aligned}
b&=-5a^2+a^2+b\\
b&=-4a^2+b\\
0&=-4a^2\\
0&=a^2\\
0&=a\end{aligned}\] The equation becomes \(y=-5x^2+0x+b\), or simply \(y=-5x^2+b\).

Since \((b,a)\) lies on the parabola, it satisfies the equation of the parabola. We can substitute \(x=b\) and \(y=a=0\) into the equation \(y=-5x^2+b\). \[\begin{aligned}
0&=-5b^2+b\\
0&=b(-5b+1)\end{aligned}\] This means that \(b=0\) or \(-5b+1=0\). Therefore, \(b=0\) or \(b=\frac{1}{5}\).

Since \(a\ \ne b\) and \(a=0\), then \(b=0\) is inadmissible.

Therefore, \(b=\frac{1}{5}\) and the equation representing the parabola \(y=-5x^2+\frac{1}{5}\). The parabola opens down and the vertex of the parabola is \(\left(0,\frac{1}{5}\right)\), and so the maximum value of the parabola is \(\frac{1}{5}\).