parabola of the form y=ax2+bx+c has a maximum value of y=4. y-coordinate of parabola at x=4 is -4 3 The y-coordinate of the parabola at x=6 is -52 3 .

Question

Determine the x-intercepts of the parabola. Enter your answer in exact form. 

smaller ( )

larger ( )

Answer 1

-4/3= a(4-h)^2 +4

-16/3 = a(4-h)^2

-52/3 = a(6-h)^2+4

-64/3=a(6-h)^2

-16/3 = a(4-h)^2

(6-h)^2/(4-h)^2= 64/16 = 4

6-h)/(4-h)= 2

6-h=8-2h

h=2

-4/3 =a(4-2)^2+4

4a= -16/3

a = -4/3

y=(-4/3)(x-2)^2 +4

=(-4/3)(x^2-4x+4)+4=0

x^2-4x+4 =-1

x^2-4x+5=0

(x-5)(x+1)=0

x intercepts = -1 and 5