Find rules for quadratic functions with graphs meeting these conditions.
1.x-intercepts at (4,0)and(-1,0) and opening upward
2.x-intercepts at (2,0)and(6,0) and maximum point=(4,12)
3.only one x-intercept and that=(-3,0) and the y-intercept=(0,18)

1.) (x - 4)(x + 1) = 0
x^2 + x - 4x - 4 = 0
y = x^2 - 3x - 4

2.) (x - 2)(x - 6) = 0
x^2 - 6x - 2x + 12 = 0
x^2 - 8x + 12 = 0
since the function has a maximum point the leading coefficient is negative.
Therefore, required equation is y = -3x^2 + 24x - 36

3.) a(x + 3)^2 = ax^2 + 6ax + 9a
since the y intercept is at y = 18 => 9a = 18 => a = 2
Therefore, required function is y = 2x^2 + 12x + 18

Ibelieve the answer is b. y=0 would make the line horizontal across the x.