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 \((2,0)\).
+33615 If the parabola has the x value of the vertex at 5, and a point on the x axis at x = 2 (which is 5-3), then it must also have a point on the x axis at (5+3,0) or at (8,0).
So you have the following two equations;
0 = a*22 + b*2 +c
0 = a*82 + b*8 + c
The slope at the vertex is zero, and the slope is given by dy/dx = 2a*x + b, so x = -b/2a. We know this x is 5 when the slope is zero, so:
5 = -b/2a
You now have three equations to find the three unknowns, a, b and c.
I'll leave the rest to you.
