Give an example of a quadratic function that has zeroes at x=2 and x=4, and that takes the value 6 when x=3..
Enter your answer in the expanded form "ax^2 + bx + c", where a,b,c are replaced by appropriate numbers.
+287 Let the function $f$ be $f(x)=ax^2+bx+c$. The zeroes of $f$ is the roots of the quadratic equation $x^2+\frac{b}{a}x+\frac{c}{a}=0$. We are given that the zeroes of $f$ are at $x=2$ and $x=4$. So the zeroes are the roots of the equation $(x-2)(x-4) = x^2-6x+8 = 0$. We are also given that $f(3)=6$. From these facts we get a system of three equations in three unknowns $a,b,c$:
\begin{eqnarray*}
\frac{b}{a} &=& -6 \\
\frac{c}{a} &=& 8 \\
9a+3b+c &=& 6.
\end{eqnarray*}
This solves to $a=-6$, $b=36$, and $c=-48$. Thus, $f(x)=-6x^2+36x-48$.
