Suppose that we have an equation $y=ax^2+bx+c$ whose graph is a parabola with vertex $(3,2)$, vertical axis of symmetry, and contains the point $(1,0)$. What is $(a, b, c)$?
The equation of a vertical parabola with a vertex of (3, 2) in vertex form is
y = A(x - 3)2 + 2
We know that (1, 0) is a solution to the equation. We can use that to find A
0 = A(1 - 3)2 + 2
0 = A( -2 )2 + 2
0 = 4A + 2
-4A = 2
A = -\(\frac12\)
So the equation of the parabola is
y = -\(\frac12\)(x - 3)2 + 2 Now we just have to get this equation into the form ax2 + bx + c
y = -\(\frac12\)(x - 3)(x - 3) + 2
y = -\(\frac12\)(x2 - 6x + 9) + 2
y = -\(\frac12\)x2 + 3x - \(\frac92\) + 2
y = -\(\frac12\)x2 + 3x - \(\frac{5}{2}\)
Now we can see that (a, b, c) = ( -\(\frac12\), 3, -\(\frac{5}{2}\) )