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How do I do this really hard math problem? Create an equation in the form f(x) = (such and so), that passes through all these points:
(2, 24) and (3, 81) and (4, 192) and (5, 575).
Please show step-by-step how to find this equation. Thanks!!!!
No mistake here. The values are from a table and we were supposed to create a function form them. Last point is indeed (5, 575). But maybe it’s not possible to create a function from these points? Hummm?
firstly u need to calculate the slope of the line
slope=(y2-y1)/(x2-x1)
=81-24/3-2
=57
now, plug in the values in the equation
y-y1=m(x-x1)
y-24=57(x-2)
y-24=57x-114
y=-(57x-11)/24 References :
Let y = f(x) , just for clarity. Makes it easier to see.
when x = 2 then y =24
when x = 3 then y= 81
when x = 4 then y =192
when x = 5 then y = 575,
hmmmmmmm
I can’t see no pattern between the numbers.
Divide each y by x we get
when x = 2 then y/2=12
when x = 3 then y/3=27
when x = 4 then y/4 =48
when x = 5 then y/5 =115
Divide by x again
when x = 2 then y/xx=6
when x = 3 then y/xx=9
when x = 4 then y/xx =12
when x = 5 then y/xx =23
hmmmm still no pattern.
seems to be a steady line where x=2, x=3 and x=4 but when x=5 it suddenly shoots off.
Its a complex one!!
But hang on… divide each by three now and we get:
when x = 2 then y/3xx=2
when x = 3 then y/3xx=3
when x = 4 then y/3xx =4
when x = 5 then y/3xx =perhaps it should be 5 and you made a typo? Would have been more obvious at the start if it were!
So it seems that x=y/(3*x*x)
multiply both sides by (3*x*x) gives
x * 3 * x * x = y
3x^3 = y
Rewriting this as per the question, f(x) = 3x^3
….and checking it out, if x = 5 then f(x)= 375 (not 575) so it seems like you or the person setting the question has done a typo or misread it!
If it’s not a typo, then it’s a more complicated curved graph. That means x=1 to x=4 cannot be a straight line or within the same curve and x=5 would then not be possible. Must be a graph with at least 3 curves in it, so f(x) will be something^3 + something^2 + something. I’m not sure if it’s meant to be really hard or if you just put that because you can’t do it!!
April 17th, 2010 at 9:45 pm
How do I do this really hard math problem?
Create an equation in the form f(x) = (such and so), that passes through all these points:
(2, 24) and (3, 81) and (4, 192) and (5, 575).
Please show step-by-step how to find this equation. Thanks!!!!
No mistake here. The values are from a table and we were supposed to create a function form them. Last point is indeed (5, 575). But maybe it’s not possible to create a function from these points? Hummm?
April 18th, 2010 at 2:47 am
3x^3 (I’m assuming you miss typed the last one where it should be 375.)
References :
April 18th, 2010 at 2:49 am
Why don’t you just plug these points into a table in your calculator. Then solve for whatever equation you need to find.
References :
April 18th, 2010 at 2:51 am
firstly u need to calculate the slope of the line
slope=(y2-y1)/(x2-x1)
=81-24/3-2
=57
now, plug in the values in the equation
y-y1=m(x-x1)
y-24=57(x-2)
y-24=57x-114
y=-(57x-11)/24
References :
April 18th, 2010 at 2:53 am
y=mx+b
(pick 2 coordinates)
192-24/4-2 = 168/2 = 84
y=84x+b
(pick 1 coordinate)
192=84(4)+b
192=336+b
-144=b
Y=84x-144
References :
my sisters brains!
April 18th, 2010 at 2:55 am
Let y = f(x) , just for clarity. Makes it easier to see.
when x = 2 then y =24
when x = 3 then y= 81
when x = 4 then y =192
when x = 5 then y = 575,
hmmmmmmm
I can’t see no pattern between the numbers.
Divide each y by x we get
when x = 2 then y/2=12
when x = 3 then y/3=27
when x = 4 then y/4 =48
when x = 5 then y/5 =115
Divide by x again
when x = 2 then y/xx=6
when x = 3 then y/xx=9
when x = 4 then y/xx =12
when x = 5 then y/xx =23
hmmmm still no pattern.
seems to be a steady line where x=2, x=3 and x=4 but when x=5 it suddenly shoots off.
Its a complex one!!
But hang on… divide each by three now and we get:
when x = 2 then y/3xx=2
when x = 3 then y/3xx=3
when x = 4 then y/3xx =4
when x = 5 then y/3xx =perhaps it should be 5 and you made a typo? Would have been more obvious at the start if it were!
So it seems that x=y/(3*x*x)
multiply both sides by (3*x*x) gives
x * 3 * x * x = y
3x^3 = y
Rewriting this as per the question, f(x) = 3x^3
….and checking it out, if x = 5 then f(x)= 375 (not 575) so it seems like you or the person setting the question has done a typo or misread it!
If it’s not a typo, then it’s a more complicated curved graph. That means x=1 to x=4 cannot be a straight line or within the same curve and x=5 would then not be possible. Must be a graph with at least 3 curves in it, so f(x) will be something^3 + something^2 + something. I’m not sure if it’s meant to be really hard or if you just put that because you can’t do it!!
References :
April 18th, 2010 at 2:57 am
WAAAAAAIT, *please* don’t tell me you mistyped that. Oh well.
Two points determine a line.
Three points determine a parabola.
Four points, then, determine a cubic.
The equation will be in the form
y = ax^3 + bx^2 + cx + d.
Plug in each set of points to get a system of equations:
24 = 8a + 4b + 2c + d;
81 = 27a + 9b + 3c + d;
192 = 64a + 16b + 4c + d;
575 = 125a + 25b + 5c + d.
Subtract the first equation from the second, and the third equation from the fourth:
57 = 19a + 5b + c;
383 = 61a + 9b + c;
Subtract your first new equation from your second:
326 = 42a + 4b.
Now solve for b:
(326 - 42a) / 4 = b;
b = 81.5 - 10.5a.
Now subtract the first original equation from the fourth, and the second from the third:
551 = 117a + 21b + 4c;
111 = 37a + 7b + c;
Multiply the second new equation by four, and then subtract it from the first new one:
107 = -31a - 7b.
Substitute in for b:
107 = -31a - 7(81.5 - 10.5a);
107 = -31a - 570.5 + 73.5a;
677.5 = 42.5a;
a ~= 15.941.
Back-substitute:
b = 81.5 - 10.5(15.941);
b ~= -85.882.
Again:
111 = 37(15.941) - (85.882)(7) + c;
c ~= 122.353.
Once more:
24 = 8(15.941) + 4(-85.882) + 2(122.353) + d;
d ~= -4.706.
Put it all together, now:
y = 15.941x^3 - 85.882x^2 + 122.353x - 4.706.
References :