Write an equation in intercept form of the parabola that passes through the point (4,−33) and has x-intercepts −7 and 5.

Answer 1

line of symmetry is x=-1 (average of the x intercepts)

vertex is (h,k) = (-1,k) = the minimum point on the parabola

this is an upward opening parabola with standard form quadratic: y-k = a(x+1)^2

plug in the point (4,-33) to get

-33 - k = 25a

k = -25a - 33

plug in the x intercept 5 or the point (5,0)

0 - 25a -33 = 36a

a = 33/11 = 3

k= -25(3)-33 = -108

y+108 = 3(x+1)^2

or

y= 3x^2 + 6x + 3 -108

y=3x^2 +6x - 105