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Find a + b + c if the graph of the equation y = ax^2 + bx + c is a parabola with vertex (5,-3), vertical axis of symmetry, and contains the point (2,8).

 Jan 12, 2021
 #1
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An equation for the parabola with its vertex at  (5 -3)  and a vertial axis of symmetry is:

 

y  =  A(x - 5)2 - 3

 

Since  (2, 8)  is a solution/point on the graph, we can plug in  2  for  x  and  8  for  y  to find  A:

 

8  =  A(2 - 5)2 - 3

 

11  =  A(2 - 5)2

 

11  =  9A

 

11/9  =  A

 

So we know that the graph of the following equation is the parabola we want:

 

\(y=\frac{11}{9}(x-5)^2-3\)

 

Now we just need to get it into the form of   y = ax2 + bx + c

 

\(y=\frac{11}{9}(x-5)(x-5)-3\\~\\ y=\frac{11}{9}(x^2-10x+25)-3\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{275}{9}-3\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{275}{9}-\frac{27}{9}\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{248}{9}\)

 

Now it is in the form we want, and so we can see that

 

a + b + c   =   \(\frac{11}{9}+-\frac{110}{9}+\frac{248}{9}\ =\ \frac{11-110+248}{9}\ =\ \frac{149}{9}\)

 

Check:  https://www.desmos.com/calculator/s4azwey3zj

 Jan 12, 2021

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