SOLUTION: I need to find the equation of the parabola that has an ordered pair (1,1), (0,2), and (3,5) as points on the graph. I need to leave the answer in the form y= ax2 + bx +c. Than

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I need to find the equation of the parabola that has an ordered pair (1,1), (0,2), and (3,5) as points on the graph. I need to leave the answer in the form y= ax2 + bx +c. Than      Log On


   



Question 32365This question is from textbook Elementary and Intermediate alegbra
: I need to find the equation of the parabola that has an ordered pair (1,1), (0,2), and (3,5) as points on the graph. I need to leave the answer in the form y= ax2 + bx +c.
Thanks
Sherri
This question is from textbook Elementary and Intermediate alegbra

Answer by mukhopadhyay(490) About Me  (Show Source):
You can put this solution on YOUR website!
The standard equation of a parabola is y = ax^2+bx+c
Substitute the ordered pairs in the above equation.
For (1,1): 1 = a+b+c.........(1)
For (0,2): 4 = c.............(2)
For (3,5): 5 = 9a+3b+c......(3)
From(2), we know c=4
From (1) and (3):
a+b = -3......(4)
9a+3b=1......(5)
Multiply (4) with -3:
-3a-3b = 9
9a+3b = 1
Sum above two equations:
6a = 10
=>a=5/3
Using equation (4): b=-3 - 5/3=-14/3
The required equation is y = (5/3)x^2-(14/3)x+4