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 $(a, b, c)$?

Guest Jun 19, 2018

#1**+2 **

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:

ElectricPavlov Jun 19, 2018