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How do you write an equation for a parabola with a vertex at -4 2 and y-intercept -2?
Wiki User
∙ 14y ago
1. The y-intercept of a parabola with equation y = ax^2 + bx + c is c. So, c = -2
2. The vertex is (x, y) = (-4, 2), where x = - b/2a
x = - b/2a
-4 = - b/2a
(-4)(-2a) = (-b/2a)(-2a)
8a = b
So we have:
y = ax^2 + bx + c (substitute what you know: 2 for y, -4 for x, 8a for b, and -2 for c)
2 = a(-4)^2 + (8a)(-4) + (-2)
2 = 16a - 32a - 2
2 = - 16a - 2 (add 2 to both sides)
4 = -16a (divide by -16 to both sides)
-1/4 = a
Since b = 8a, then b = 8(-1/4) = -2 Since a = -1/4, b = -2, and c = -2, then we can write the equation of the parabola as
y = (-1/4)x^2 - 2x - 2.
Wiki User
∙ 14y ago
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