SOLUTION: What is the equation of a parabola with its axis of symmetry at line of x=3/4, y-intercept at the point (0,1), and passing through the point of (2,3)?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the equation of a parabola with its axis of symmetry at line of x=3/4, y-intercept at the point (0,1), and passing through the point of (2,3)?      Log On


   

Question 696155: What is the equation of a parabola with its axis of symmetry at line of x=3/4, y-intercept at the point (0,1), and passing through the point of (2,3)?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
y=a%28x-3%2F4%29%5E2%2Bk is the equation of a parabola with axis of symmetry x=3%2F4
If we substitute the coordinates of point (0,1), we find
1=a%280-3%2F4%29%5E2%2Bk-->1=9a%2F16%2Bk-->1-k=9a%2F16-->16-16k=9a-->9a%2B16k=16
If we substitute the coordinates of point (2,3), we find
3=a%282-3%2F4%29%5E2%2Bk-->3=a%285%2F4%29%5E2%2Bk-->3=25a%2F16%2Bk-->3-k=25a%2F16-->48-16k=25a-->25a%2B16k=48
We now have the system
system%2825a%2B16k=48%2C9a%2B16k=16%29
Subtracting the second equation from the first we get
25a%2B16k-%289a%2B16k%29=48-16-->25a%2B16k-9a-16k=48-16-->16a=32-->highlight%28a=2%29
Substituting that value back into 9a%2B16k=16 we get
9%2A2%2B16k=16-->18%2B16k=16-->16k=16-18-->16k=-2-->highlight%28k=-1%2F8%29
So the equation is
highlight%28y=2%28x-3%2F4%29%5E2-1%2F8%29
It can be transformed, like this:
y=2%28x%5E2-%283%2F2%29x%2B9%2F16%29-1%2F8-->y=2x%5E2-3x%2B9%2F8-1%2F8-->highlight%28y=2x%5E2-3x%2B1%29