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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.

Jun 9, 2021

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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, $f(x)=-6x^2+36x-48$.

Jun 12, 2021