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.

P (x) = a (x - 2) (x - 4) and we need to find "a"

And when x = 3 we have that

a( 3 - 2) (3 - 4) = 6 simplify and solve for "a"

a (-1) = 6 → a = -6

So we have

-6 [ x^2 - 6x + 8] =

P(x) = -6x^2 + 36x - 48