|
Question 662682: Hye!please help me on this Q..
find the eqn of parabola that has a horizontal axis and passes through the points (-1,1), (11,-2), (5,-1).
thanks!!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the eqn of parabola that has a horizontal axis and passes through the points (-1,1), (11,-2), (5,-1).
**
y=Ax^2+Bx+C
..
1=A(-1)^2+B(-1)+C
-2=A(11)^2+B(11)+C
-1=A(5)^2+B(5)+C
..
1=A-B+C
-2=121A+11B+C
-1=25A+5B+C
..
1=A-B+C
-2=121A+11B+C
subtract
3=-120A-12B
..
1=A-B+C
-1=25A+5B+C
subtract
2=-24A-6B
..
3=-120A-12B
4=-48A-12B
subtract
-1=-72A
A=1/72
6B=-24A-2=-1/3-6/3=-7/3
B=-7/18
C=B-A+1
C=-7/18-1/72+1
C=-28/72-1/72+72/72
C=43/72
A=1/72, B=-7/18, C=43/72
Check:
121A+11B+C
=121/72-77/18+43/72
=121/72-308/72+43/72
=-144/2
=-2
Equation of parabola:
y=x^2/72-7x/18+43/72
note: this parabola has a vertical axis of symmetry; whereas, problem stated parabola has a horizontal axis of symmetry
|
|
|
| |
