The equation of the parabola could be written as y-k = a(x-h)^2, where (h,k) is the vertex. Thus, y-(-3) = a(x+4)^2, or y+3 = a(x+4)^2.
The coordinates of one x-intercept are (-11,0). Thus, y+3 = a(x+4)^2 becomes 0+3 = a(-11+4)^2, so that 3 = a(-7)^2, or 3 = 49a. Therefore, a = 3/49, and the equation of the parabola becomes
y+3 = (3/49)(x+4)^2.
To find the other x-intercept, let y = 0 and solve the resulting equation for x:
0+3 = (3/49)(x+4)^2, or (49/3)*2 = (x+4)^2
Taking the sqrt of both sides, plus or minus 49/3 = x+4.
plus 49/3 = x+4 results in 37/3 = x, whereas
minus 49/3 = x+4 results in x = -61/3. Unfortunatelyi, this disagrees with what we are told: that one x-intercept is x= -11, or (-11,0).
Trying again, using the quadratic equation y=ax^2 + bx + c, we substitute the coordinates of the points (-4,-3) and (-11,0) and solve for {a, b, c}:
-3 = a(-4)^2 + b(-4) + c, or -3 = 16a - 4b + c
0 = a(-11)^2 - 11b + c, or 0 = 121a - 11b + c
If the vertex is at (-4,-3), then, because x= -b/(2a) also represents the x-coordinate of the vertex, -4 = b / (2a), or -8a = b, or 0 = 8a + b
Now we have 3 equations in 3 unknowns:
0 = 8a + 1b -3 = 16a - 4b + c 0 = 121a - 11b + c
This system of 3 linear equations can be solved in various ways. I've used matrices, finding that a, b and c are all zero. This is wrong.
So, let's try again. Recall that x = -b / (2a) is the axis of symmetry, which in this case is x = -4. If one zero is at -11, this point is 7 units to the left of x = -4. The other zero is 7 units to the right of x = -4, that is, at x = 3.
Now we have 3 points on the parabola: (-11,0), (-4,-3) and (3,0).
This is sufficient info for us to determine {a,b,c} in y=ax^2+bx+c. One by one we take these 3 points and subst. their coordinates into y=ax^2+bx+c, obtaining 3 linear equations:
