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

Question

Ethan Adkins · Accepted Answer

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

Kennedy Adams · Answer

undefined To find the equation of the parabola, you can use the standard form equation $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. Given the axis of symmetry is $x = 3$, the vertex's x-coordinate is 3. Using the given points $(1,0)$ and $(-4,3)$, you can find the y-coordinate of the vertex. Then, substitute the vertex coordinates and one of the given points into the equation to solve for $a$. Finally, plug in the values of $a$, $h$, and $k$ into the standard form equation to get the equation of the parabola.