Question

I have tried several times to solve this. I have 3 points (0,-5), (3, -6), (-3,-7). I'm trying to get the quadratic equation. I keep getting y=(-1/6)x^2+ (1/6)x +0. When i plug in other numbers to check it's not the parabola facing down like my image. Please help!

Answer 1

let y = ax^2 + bx + c
for (0,-5)
-5 = 0 + 0 + c, so c = -5
for (3,-6)
-6 = 9a + 3b - 5
<b>9a + 3b = -1</b> , #1
for (-3,-7)
9a -3b -5 = -7
<b>9a - 3b = -2</b> , #2
add #1 and #2
18a = -2
a = -1/+
sub into #1
9(-1/6) + 3b = -1
-3/2 + 3b = -1
-3 + 6b = -2
6b = 1
b = 1/6

so f(x) = (-1/6)x^2 + (1/6)x - 5

All 3 points satisfy this equation, you have no constant in your equation

You seem to have made the error in finding the constant (which was the easiest part)

Answer 2

THANK YOU! THANK YOU! THANK YOU!!!