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I have a given set of values for x and y and am supposed to find the equasion that works for all of them, but there is a catch one of the y values is incorrect (It needs to be increased by one) x=3 y= -4.5, x=5 y=4.5, x=6 y =10, x=8 y= 24, x=4 y=0 Are the values I am given. There is supposedly one equation which can satisfy all of these, (when the y value is correct) but I have been very unsuccessful in figuring out which y variable is inncorrect and the equasion, any help would be very appreciated.

2006-10-09 18:49:02 · 4 answers · asked by Matt K 1 in Education & Reference Homework Help

4 answers

I solved it this morning.

It is a quadratic. The point that is incorrect is the first one.
use x=3 y=-3.5

set up
y=Ax^2 +Bx + c

using x=4, y=0
0=16A + 4B +c

using x=6, y=10
10=36A +6B +C

Subtract the second from the first:
10=20A - 2B

using x=5, y=4.5
4.5=25A + 5B + C

Using x= 3 y=-3.5

-3.5 =9A + 3B + C

Subtract the second from the first

8 = 16A + 2B

Now add 10=20A - 2B

to get
18=36A
A=.5

Plug this into one of the equations above to get
b=0

Now we know that
y=.5x^2 + C

using x=4, y=0
0=8 + C
c = -8

the final solution is

y=.5x^2 -8

It works for all of the points when you use
x=3, y=-3.5

j

2006-10-09 19:30:00 · answer #1 · answered by odu83 7 · 0 0

am getting 2 y values as incorrect
well the equation i got is
9x-2y=36

now substitute the first point (3,-4.5) it works
we get 27+9=36

then second (5,4.5)
we get 45-9=36

then (6,10) 54-20<>36
then (8,24) 72-48<>36
(4,0) 36-0 it works..

i some how feel the values u gave are wrong.

cos lets take 2 points among those u gave
(6,10) (4,0) now there can be situtaion one of them is the wrong one or none.

case 1
lets say (6,10) is wrong then the correct point should be (6,11) so now we get an equation using these2 points that is
11x - 2y = 44 you may verify it with substituing the points

now this equation which we got should work out with the remaning points (3,-4.5) (5,4.5) (8,24) but it doesnt work out.
so our assumption is wrong.i.e (6,10) is correct point

case 2
now say (4,0) is wrong then correct point should be (4,1) so
now we get an equation using these2 points that is
9x - 2y = 34

now this equation which we got should work out with the remaning points (3,-4.5) (5,4.5) (8,24) but it doesnt work out.
so our assumption is wrong.i.e (4,0) is correct point

case 3
now if both (6,10) and (4,0) are correct points then the equation with them should work for atleast 2 points in the remaning.
the equation is 5x-y=20
now checking for each point in (3,-4.5) (5,4.5) (8,24)
(3,-4.5) 15+4.5<> 20
(5,4.5) 25 -4.5<> 20
(8,24) 40-24<>20

so this shows clearly that the given question is wrong..:)

2006-10-09 19:55:12 · answer #2 · answered by just08in 2 · 0 0

The only things that works is that there are 2 y's that are incorrect. 10 should be 9 and 24 should be 18. y=4.5x-18 then.

2006-10-09 19:21:21 · answer #3 · answered by Anonymous · 0 0

I think that maybe you might be able to find a better answer searching on yahoo, becuase I have yet to get a substantial answer in here.

best of luck to you.

2006-10-09 19:57:23 · answer #4 · answered by Anonymous · 0 0

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