A parabola y=ax^2+bx+c has a line of symmetry x=-2 and goes through the two points (5,1) and (4,-3). Find a + b + c.
If the line of symmetry goes through x = -2.....then this is the x coordinate of the vertex
So we have this
y = a ( x - h) + k where h = -2
We need to solve these two equations for a and c
1 = a ( 5 + 2)^2 + k → 1 = 49a + k → 49a + k = 1 (1)
-3 = a(4 + 2)^2 + k → -3 = 36a + k → 36 a + k = -3 (2)
Subtract (2) from (1) and we have that
13a = 4
a = 4/13
And
49(4/13) + k = 1
k = 1 - 196/13
k = -183/13
So we have
y = (4/13) ( x + 2)^2 - 183/13
13 y = 4 (x^2 + 4x + 4) -183
13y = 4x^2 + 16x + 16 - 183
13y = 4x^2 + 16x - 167
y = (4/13)x^2 + (16/13)x - 167/13
So
a + b + c = ( 4 + 16 - 167) /13 = -147/13