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 $(a, b, c)$?
+36915 Let's write the beginning of the parabola in vertex form
y= a(x-h)^2 +k (h,k) is the vertex ...substitute
y= a(x-3)^2 + 2 expand
y = a (x^2 - 6x +9) + 2 now substitute the point (1,0) in to the equation
0 = a (1^2 -6 +9) + 2
0 = 4a + 2
so a = -1/2 in THIS equation and it looks like this
y = -1/2 (x^2-6x+9) +2 Simplify
y = -1/2x^2 +3x -4.5 + 2
y= -1/2 x^2 + 3x - 2.5
Here is a graph:
