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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 $(a, b, c)$?

 Jun 30, 2019
 #1
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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,  -\(\frac{5}{2}\) )

 

Check: https://www.desmos.com/calculator/jwb2hcg1yy

 Jun 30, 2019

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