Find the equation for a parabola with vertex at (7,2) and a y-intercept of (0,16)

Answer 1

note that (0,16) means f(0)=16 which forces the constant term C=16

vertex happens at -b/(2a) = 7

-b = 14a

b = -14a

So the quadratic function is:

f(x) = ax^2- (14a)x + 16

(7,2) means f(7)=2:

2 = 49a - 98a + 16

-14 = -49a

-14/-49 = a

a = 2/7 ----> b = -4

f(x) = (2/7)x^2 -4x + 16