SOLUTION: 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. I know that this parabola will be a ma

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 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. I know that this parabola will be a ma      Log On


   

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. I know that this parabola will be a maximum, and the y-intercept is at 1.8.
I have tried putting the equations into each other by plugging in each point to their own equation, and then solving from that. I must have gone wrong somewhere because I ended up with y= 0.23x^2 - 0.64x + 1.8. I don't know how to move on, could you please help?
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Rethink the points which are given. A vertex maximum should occur between x=0 and x=4, and thinking visually or plotting on paper, you should understand and expect because of this maximum vertex, coefficient on x^2 will be a negative value.

Not seeing any more convenient method, form a system of three equations using the points.

y=ax^2+bx+c

system%28a%2A0%5E2%2Bb%2A0%2Bc=1.8%2Ca%2A2%5E2%2Bb%2A2%2Bc=4%2Ca%2A4%5E2%2Bb%2A4%2Bc=3%29

system%280%2Aa%2B0b%2Bc=1.8%2C4a%2B2b%2Bc=4%2C16a%2B4b%2Bc=3%29

Just substitute for c and obtain a system of two equations in a and b.
system%284a%2B2b%2B1.8=4%2C16a%2B4b%2B1.8=3%29
-
system%284a%2B2b=2.2%2C16a%2B4b=1.2%29

Preparing for start of Elimination Method,
system%288a%2B4b=4.4%2C16a%2B4b=1.2%29
SUBTRACT the first equation from the second equation,
8a%2B0b-3.2
8a=-3.2
a=-3.2%2F8, and multiply by 5, just to simplify into a rational number.
a=-16%2F40
a=-4%2F10
highlight%28a=-2%2F5%29 and already was found highlight%28c=1.8%29.

Pick either equation from the two-equation, two variable system to solve for b.
4a%2B2b=2.2
2b=2.2-4a
b=1.1-2a
b=1.1-2%28-2%2F5%29
b=1.1%2B4%2F5
b=1.1%2B0.8
highlight%28b=1.9%29

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
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.
***
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