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.
Why are you appologising for the Latex, why didn't you just edit it out?
Since it has zeroes at \(x = 2\) and \(x = 4\), the quadratic can be written in the form \(y = a(x-2)(x-4)\). You can substitute the point \((3,6)\) to get that \(6=a(3-2)(3-4)\), so \(a=-6\). Thus, we get the answer of \(-6(x-2)(x-4) = -6(x^2-6x+8) = -6x^2 +36x-48\).
hope this helped