Create a quadratic equation where the parabola has a maximum vertex and has a y intercept at (0,7)

Answer 1

for the vertex to be a maximum point, the coefficeint of the x^2 term must be negative

y =a(x-h)^2 + k where a<0 and (h,k) = vertex

when x=0, y= the intercept = 7

7 = a(h)^2 + k

let (h,k) = (2,11)

then

7 =4a+11

a=-1

y=-(x-2)^2+11

or

y=-x^2+4x+7