How do you find the equation for a parabola with an axis of symmetry of x=3 and points (1,0) and (-4,3)?

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Answer 1 · 2015-05-10T01:49:28

Equation #y = ax^2 + bx + c#. Find a, b, and c.

axis of symmetry: #-b/(2a) = 3 -> b = -6a## (1)

y passes at point (1, 0) and point (-4, 3)

0 = a + b + c (2) -> c = - a - b = - a + 6a = 5a

3 = 16a - 4b + c (3) -> 3 = 16a + 4(6a) + 5a = 16a + 24a + 5a = 45a

45a = 3 --> a = 3/45 = 1/15
b = -6/15
c = 5/15
Equation #f(x) = x^2/15 -(6x)/15 +5/15. #

Check:
#x = 1 --> y = 1/15 - 6/15 + 5/15 = 0# Correct
#x = -4 -> y = 16/15 + 24/15 + 5/15 = 45/15 = 3# . Correct