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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
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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

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