determine quadratic function

Question

minimum value of 12 at x=-4 and y-intercept of 60 determine quadratic function

Answer 1

Vertex at (-4,12) so quadratic can be written:

y-12=a(x+4)², where constant a has to be found.

When x=0, y=60 (intercept):

60-12=16a, 16a=48, a=3.

y-12=3(x+4)²=3(x²+8x+16)=3x²+24x+48,

y=3x²+24x+60 is the quadratic.