Write an equation of a parabola with​ x-intercepts at (1/2,0) and (-8,0) which passes through the point (0,8).

Answer 1

zeros are -8 and 1/2

factors are (x+8)(x-1/2)

y = a(x+8)(x-1/2) plug in the point (0,8)

8 = a(8)(-1/2) = -4a

a= -2

y = -2(x+8)(x-1/2)

= (x+8)(1-2x)= -2x^2 -15x +8

y = -2x^2 -15x + 8

or

complete the square

y = -2(x^2+15/2x) + 8

y = -2(x^2 +15/2x+ 225/4) + 8+ 225/2

y = -2(x+15/4)^2 + 241/2

y = -2(x+ 3 3/4)^2 + 120 1/2

in vertex form: y=a(x-h)^2+k, where (h,k)=

vertex = (-3 3/4, 120 1/2)= maximum point of the downward opening parabola

Answer 2

y = a(x – ½)(x + 8) is the general equation for parabolas with​ x-intercepts at (½,0) and (-8,0).

Only one of these parabolas will pass through the point (0,8).

y = a(x – ½)(x + 8) = a(x² + 7.5x – 4)

To find the equation for the parabola that passes through (0,8):

plug in x = 0 and y = 8, then solve for a.

8 = a(0² + 7.5(0) – 4) = –4a

a = –2

y = –2(x² + 7.5x – 4) = –2x² – 15x + 8

Answer

y = –2x² – 15x + 8