Determine the general equation of a quartic function with end behaviour \(f(x) \rightarrow -\infty\) as \(x \rightarrow \pm\infty\), a zero at \(x=\frac{1}{2}\), and another at \(x=4\) of multiplicity \(3\).
Solution
Factor to produce the zero:
\(\left(x-\frac{1}{2}\right)\) or \((2x-1)\)
\(\begin{align*}x&=\frac{1}{2}\\2x&=1\\2x-1&=0\end{align*}\)
\(\begin{align*}x&=4\\x-4&=0\end{align*}\)
Therefore, the factors \((2x-1)\) and \((x-4)\) will produce the necessary zeros. So,
\[\begin{align*}y&=k(2x-1)(x-4)(x-4)(x-4)\\&=k(2x-1)(x-4)^3\end{align*}\]
The information about the end behaviour tells us the graph of the function opens downward.
This implies a negative leading coefficient.
The general equation for the family of functions with the given zeros would be
\[y=k(2x-1)(x-4)^3, k\lt 0, k \in \mathbb{R}\]