I NEED YOUR HELP ASAP!! Grade 11 applied math question in quadratics!!!

Question

Question: a parabola has a vertex of (4,8) and opens downward. Write two possible equations of quadratic functions in vertex form such that the parabola has exactly two zeros. Find the zeroes for both functions.

Answer 1

y = f(x) = -(x-4)^2 + 8 has vertex (4,8) and opens downward

0 = -(x-4)^2 + 8

(x-4)^2 = 8

x+4 = + or - sqr8

x= -4+sqr8 or -4 -sqr8

those are the two zeros

another parabola with same vertex opening downward

is

y= f(x) = -2(x-4)^2 + 8

0 =-2(x-4)^2 + 8

(x+4)^2 = 4

x+4 = + or -2

x=-4 + or -2 = 6 or 2

zeros are when y=0, the place the parabola intersects the x axis

any quadratic equation solved for y with a negative coefficient of the x^2 term will be downward opening

y=a(x-h)^2 + k will be downward opening if a<0. Its vertex is (h,k)

set y =0 and solve for x, the 2 solutions are 2 zeros, although some solutions are imaginary or just one with multiplicity 2. The just one solution would be when the vertex had a y coordinate of 0