y= ax^2 + bx +c

a) touches the x-axis at 4 and passes through (2,12)

touches the x-axis at 4 means that passes trough (4,0) and b^2 - 4*a*c = 0 (the quadratic has 1 solution)

- passes trough (4,0) that is for x=4 y=0

4 = a*0^2 + b*0 + c

c = 4

passes through (2,12) that is x = 2 and y = 12

12 = a*2^2 + b*2 + c

4*a + 2*b + c = 12

since c = 4

4*a + 2*b = 8

2a + b = 4

and b^2 - 4*a*c = 0 that is

b^2 - 16*a = 0

by solving the system of equations

2a + b = 4

b^2 - 16*a = 0

we find

a = 4 + 3^0.5 and b = - 4 - 2*(3*^0.5)

a = 4 - 3^0.5 and b = - 4 + 2*3^0.5

the equation of the quadratic is

for a = 4 + 3^0.5, b = - 4 - 6^0.5 and c = 4

y= ax^2 + bx +c = (4 + 3^0.5)*x^2 + (4 - 2*(3^0.5))*x + 4

for a = 4 - 3^0.5, b = - 4 + 2*3^0.5 and c = 4

y= ax^2 + bx +c = (4 - 3^0.5)*x^2 + (4 + 2*(3^0.5))*x + 4

b) y= ax^2 + bx +c has vertex (-4,1) and passes through (1,11)

1 = 16a - 4b + c

11 = a + b + c

the vertex is x = -b/2a that is -b/2a = -4

by solving the system of equations

16a - 4b + c = 1

a + b + c = 11

-b/2a = -4

we find

a = 0.4

b = 3.2

c = 7.4

the equation of the quadratic will be

y = 0,4x^2 + 3.2x + 7.4