### Instruction: 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 $(a, b, c)$?### Response: So first, we know that the vertex form of a parabola is $a(x-h)^2 + k$.
And we know that the vertex is $(3, 2)$, so $(h, k) = (3, 2)$.
So now we have $a(x-3)^2 + 2$.
And if we plug in $(1, 0)$ into the equation, we get $a(1-3)^2 + 2 = 0$.
We can simplify that to $a(4) + 2 = 0$.
So, $4a = -2$, which means $a = -2/4 = -1/2$.
And now we can plug that back into the equation to get $-1/2(x-3)^2 + 2$.
If we expand that, we get $-1/2(x^2 - 6x + 9) + 2$.
Simplifying, we get $-1/2x^2 + 3x - 9/2 + 2$.
And that simplifies to $-1/2x^2 + 3x -5/2$.