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# parabola

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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+0 Answers

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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   =   $$\frac{11}{9}+-\frac{110}{9}+\frac{248}{9}\ =\ \frac{11-110+248}{9}\ =\ \frac{149}{9}$$

Jan 12, 2021