Which function has a vertex at (2, –9)? f(x) = –(x – 3)2 f(x) = (x + 8)2 f(x) = (x – 5)(x + 1) f(x) = –(x – 1)(x – 5)

The solutions or roots so-called are really just the x-intercepts, and for a quadratic equation, the vertex is found right half-way between those x-intercepts.

we can get the intercepts by zeroing out "y" or f(x), so let's take a peek at one of those,



so, the x-intercepts or solutions are at 5 and -1, let's take a peek what's the halfway point.

-1----0----1----  2  ----3-----4------5

well then, now we know the vertex is at x = 2, but, what's the y-coordinate of it anyway?

y = (x - 5)(x + 1)

y = (2 - 5)(2 + 1)

y = (-3)(3)

y = -9

C. i believe it is (5×(-20))+(100×(0.7))