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).
+9466 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 = \(\frac{11}{9}+-\frac{110}{9}+\frac{248}{9}\ =\ \frac{11-110+248}{9}\ =\ \frac{149}{9}\)
