Write the equation for the parabola that has x− intercepts (−2,0) and (1.2,0) and y− intercept (0,−4)

Answer 1

y = a (x - h)^2 + k ... (h,k) is the vertex

h is midway between the x-intercepts (roots) ... (-2 + 1.2) / 2 = -.4

from (-2,0) ... 0 = a (-2 - -.4)^2 + k ... 0 = 2.56 a + k

from (0,-4) ... -4 = a (0 - -.4)^2 + k ... -4 = .16 a + k

solve the system for a and k

subtracting equations (to eliminate k) ... 4 = 2.4 a ... a = 5/3

substituting ... -4 = (4/25 * 5/3) + k = 4/15 k ... k = -64/15

y = 5/3 (x + 2/5)^2 - 64/15

Answer 2

x− intercepts (−2,0) and (1.2,0) means that
y = a(x+2)(5x-6)
Now plug in (0,-4) to find a:
a(2)(-6) = -4
-12a = -4
a = 1/3
y = 1/3 (x+2)(5x-6) = 1/3 (5x^2 + 4x - 12)