so we have x-intercepts of 0 and 3
thus
f(x) = kx(x-3)
but (5,30) also lies on f(x), so ...
30 = k(5)(2)
10k = 30
k = 3
f(x) = 3x(x-3) or f(x) = 3x^2 - 9x
Find a possible formula for a polynomial f with the following properties.
f has degree less-than-or-equal-to 2, f(0) = f(3) = 0 and f(5) = 30
f(x) = ?
k(x+0)(x+0)(x-3)
I know the zeroes for 3 of them are 0, 0, 3, how would i find out the rest?
1 answer
1 answer