Question 919274
I am given three points on a graph (0, 1.8) (2,4) and (4,3) and am asked to find the equation of the parabola that these points pass through.
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Standard form of equation for a parabola:
y=Ax^2+Bx+c
For given points on parabola:
(0,1.8) 1.8=0A+0B+C
(2,4)   4=4A+2B+C
(4,3)   3=16A+4B+C
..
C=1.8 (fm 1st eq)
4=4A+2B+1.8
3=16A+4B+1.8
..
4A+2B=2.2
16A+4B=1.2
..
16A+8B=8.8 (mult. eq by4)
16A+4B=1.2
subtract:
4B=7.6
B=1.9
4A=2.2-2B2.2-3.8=-1.6
A=-0.4
..
A=-0.4
B=1.9
C=1.8
Equation of parabola: y=-0.4x^2+1.9x+1.8
..
Check:
4A+2B+C=4*(-0.4)+2*1.9+1.8=4
16A+4B+C=16*(-0.4)+4*1.9+1.8=3