The vertex of a parabola is at (12,-4), and its axis of symmetry is vertical. One of the x-intercepts is at (-15,0). What is the x-coordinate of the other x-intercept?
+186 The equation for a parabola is y=ax^2+bx+c. We are given two points, and the axis of symmetry (x=12) which will be plenty for us.
The axis of symmetry can be found with the equation -b/2a=12. Solving for b, we get -b=24a, so b=-24a. Now we can turn our equation into:
y=ax^2-24ax+c. We can plug in the two points to get the two equations
-4=-144a+c
0=585a+c
Solving the system of linear equations, a=4/729 and c=-260/81. Now we can solve for b=-96/729.
Finding the other root of the quadratic 4/729x^2-96/729x-260/81=0, we get the other x-intercept to be (39, 0) or 39.
