Question 889601
find the equation of the parabola with vertex on the line y=x, the axis of parabola is parallel to the x-axis and passing through (6,-2)and (3,4)
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give data shows parabola open rightwards:
Its vertex form of equation: {{{x=(y-k)^2+h}}}
(6,-2)   6=(-2-k)^2+h
(3,4)    3=(4-k)^2+h
..
6=4+4k+k^2+h
3=16-8k+k^2+h
subtract
3=-12+12k
12k=15
k=15/12=5/4
..
3=(4-k)^2+h
sub k
3=(4-5/4)^2+h
3=(11/4)^2+h
48/16=121/16+h
h=-121/16+48/16=-73/16
equation:{{{x=(y-5/4)^2-73/16}}}