.
Find the equation of the parabola with axis parallel with the x-axis
and passing through (4,2), (-4,10) and (-4, -6). Sketch the graph
~~~~~~~~~~~~~~~~
This problem is not to make stupid calculations (as other tutor suggests).
It is to develop your observation skills and mind, as well as the sense of beauty.
Notice that the points (-4,10) and (-4,-6) have the same x-coordinate.
Hence, the horizontal axis of the parabola is the perpendicular bisector
y= = 2
of the segment, connecting (-4,10) and (-4,-6).
Also notice that the point (4,2) is the vertex, since it lies on the axis y= 2.
So, the vertex form of this parabola is x = a*(y-2)^2 + 4. (*)
The coefficient "a" is unknown now.
Find it from the condition that the point (-4,10) lies on the parabola.
Substitute x=-4, y= 10 into equation (*). You will get
-4 = a*(10-2)^2 + 4
-4 - 4 = 64a, ---> 64a = -8 ---> a = = .
So, the equation of the parabola is x = .
Solved.