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

May 10, 2015

Equation $y = a {x}^{2} + b x + c$. Find a, b, and c.

axis of symmetry: $- \frac{b}{2 a} = 3 \to b = - 6 a$# (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 $f \left(x\right) = {x}^{2} / 15 - \frac{6 x}{15} + \frac{5}{15.}$

Check:
$x = 1 - \to y = \frac{1}{15} - \frac{6}{15} + \frac{5}{15} = 0$ Correct
$x = - 4 \to y = \frac{16}{15} + \frac{24}{15} + \frac{5}{15} = \frac{45}{15} = 3$ . Correct