This seems pretty simple and straightforward, but I can't seem to figure it out.
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.
Thank you very much for your help!
Zeroes are at x = 2 and x = 4 \(\Longleftrightarrow\) \(f(x) = k(x - 2)(x - 4)\)
In this case, f(3) needs to be equal to 6.
\(f(3) = k(3 - 2)(3 - 4) = -k\)
So -k = 6, and k must be -6.
\(f(x) = -6(x - 2)(x - 4)\)
You can finish it from here.