Is y = -3x^2 - 4x + 3 supposed to be your answer ?
Can't be right, since (0,-4) clearly does not satisfy your equation.
Method1:
let the equation be ax^2 + bx+c = y
for (0,-4), 0+0+c=-4 , c = -4
for (-2,-20), 4a-2b - 4 = -20
4a - 2b = -16
2a - b = -8
for (4,-20), 16a + 4b - 4 = -20
4a + b = -4
add the last two equations:
6a = -12
a = -2
back in 2a-b=-8
-4-b=-8
b = 4
so y = -2x^2 + 4x - 4
Method2:
the x of the vertex must lie half-way between -2 and 4 which would be 1
so we must have y = a(x-1)^2 + q
for (0,-4) ---> -4 = a+q
for (4,-20) --> -20 = 9a + q
subtract them:
-16 = 8a = -2
sub into a+q=-4
-2+q = -4
q = -2
equation: y = -2(x-1)^2 - 2
expand, if you must, to get the same equation as in Method 1
Find a quadratic model for the set of values: (-2,-20), (0,-4), (4,-20). Show your work.
y = -3x^2 - 4x + 3
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