Quadratic functions

Question

determine the standard form quadratic function, given the following: one of the x-intercepts at x = 6, and with vertex (4,2)

Answer 1

the x intercepts are equidistant from the axis of symmetry which is x=4

if x=6 is one intercept, x= 2 is the other. 6-4 = 2 = 4-2

y = a(x-4)^2 + 2

plug in either x intercept (2,0) or (6,0) to solve for a

0= a(6-4)^2 +2

-2 = a(2)^2

a = -2/4 = -1/2

the parabola is y=-(1/2)(x-4)^2 + 2

this equation works for plugging in the vertex (4,2) and both x intercepts (6,0) & (2,0)

2= -(1/2)(0)^2 + 2 = 2

0 = -(1/2)(2-4)^2 + 2= -2+2= 0

0= -(1/2)(6-4)^2 +2 = -2 +2 =0

in standard y =ax^2 + bx + c form

y=-(1/2)(x-4)^2 + 2

2y = -(x^2 -8x +16) +4

2y = -x^2 +8x -16 +4

2y = - x^2 + 8x -12

y = (-1/2)x^2 + 4x -6

where a =-1/2 b = 4 and c=-6