Writing a rational function representing Average cost over time. (Pre-Calc)
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1
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I have a linear function that is C(t) = 4999.99 + 36(t)
where C(t) = Cost @ year after purchase
And t = years since purchase
-
AC(t) = (4,999.99 +36(t)) / (t)
(This is my funciton)
-
But, acording to my projects description. AC(t) (average cost over time is AC(t) = C(t) / t
but every time i graph it i get something that looks like this
But i know it needs to look something like a curve.
Im confused since the projects description says to do this:
⦁ The average cost, in dollars per year, AC(t) , will turn out to be the total cost divided by . AC(t) is a rational function. Write the rational function representing the average cost function per year for your device.
functions rational-functions
asked Nov 16 at 2:57
xannax159
83
|
show 2 more comments
up vote
1
down vote
favorite
I have a linear function that is C(t) = 4999.99 + 36(t)
where C(t) = Cost @ year after purchase
And t = years since purchase
-
AC(t) = (4,999.99 +36(t)) / (t)
(This is my funciton)
-
But, acording to my projects description. AC(t) (average cost over time is AC(t) = C(t) / t
but every time i graph it i get something that looks like this
But i know it needs to look something like a curve.
Im confused since the projects description says to do this:
⦁ The average cost, in dollars per year, AC(t) , will turn out to be the total cost divided by . AC(t) is a rational function. Write the rational function representing the average cost function per year for your device.
functions rational-functions
asked Nov 16 at 2:57
xannax159
83
I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a linear function that is C(t) = 4999.99 + 36(t)
where C(t) = Cost @ year after purchase
And t = years since purchase
-
AC(t) = (4,999.99 +36(t)) / (t)
(This is my funciton)
-
But, acording to my projects description. AC(t) (average cost over time is AC(t) = C(t) / t
but every time i graph it i get something that looks like this
But i know it needs to look something like a curve.
Im confused since the projects description says to do this:
⦁ The average cost, in dollars per year, AC(t) , will turn out to be the total cost divided by . AC(t) is a rational function. Write the rational function representing the average cost function per year for your device.
functions rational-functions
asked Nov 16 at 2:57
xannax159
83
I have a linear function that is C(t) = 4999.99 + 36(t)
where C(t) = Cost @ year after purchase
And t = years since purchase
-
AC(t) = (4,999.99 +36(t)) / (t)
(This is my funciton)
-
But, acording to my projects description. AC(t) (average cost over time is AC(t) = C(t) / t
but every time i graph it i get something that looks like this
But i know it needs to look something like a curve.
Im confused since the projects description says to do this:
⦁ The average cost, in dollars per year, AC(t) , will turn out to be the total cost divided by . AC(t) is a rational function. Write the rational function representing the average cost function per year for your device.
functions rational-functions
functions rational-functions
asked Nov 16 at 2:57
xannax159
83
asked Nov 16 at 2:57
xannax159
83
asked Nov 16 at 2:57
xannax159
83
asked Nov 16 at 2:57
xannax159
83
asked Nov 16 at 2:57
xannax159
83
83
I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13
|
show 2 more comments
I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13
I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You are doing fine. The total cost is $5000$ at the start and $36$ per year thereafter. The average cost starts very high and decreases as the $5000$ gets averaged over more years. It would be better to plot your curve over a range like $0-10$ or $0-30$, not up into the hundreds. You can write the average cost as $frac {5000}t+36$ and you can see that at $t=1$ it will be $5036$, at $t=2$ it will be $2518$ and so on.
answered Nov 16 at 3:15
Ross Millikan
288k23195365
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are doing fine. The total cost is $5000$ at the start and $36$ per year thereafter. The average cost starts very high and decreases as the $5000$ gets averaged over more years. It would be better to plot your curve over a range like $0-10$ or $0-30$, not up into the hundreds. You can write the average cost as $frac {5000}t+36$ and you can see that at $t=1$ it will be $5036$, at $t=2$ it will be $2518$ and so on.
answered Nov 16 at 3:15
Ross Millikan
288k23195365
add a comment |
up vote
1
down vote
accepted
You are doing fine. The total cost is $5000$ at the start and $36$ per year thereafter. The average cost starts very high and decreases as the $5000$ gets averaged over more years. It would be better to plot your curve over a range like $0-10$ or $0-30$, not up into the hundreds. You can write the average cost as $frac {5000}t+36$ and you can see that at $t=1$ it will be $5036$, at $t=2$ it will be $2518$ and so on.
answered Nov 16 at 3:15
Ross Millikan
288k23195365
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are doing fine. The total cost is $5000$ at the start and $36$ per year thereafter. The average cost starts very high and decreases as the $5000$ gets averaged over more years. It would be better to plot your curve over a range like $0-10$ or $0-30$, not up into the hundreds. You can write the average cost as $frac {5000}t+36$ and you can see that at $t=1$ it will be $5036$, at $t=2$ it will be $2518$ and so on.
answered Nov 16 at 3:15
Ross Millikan
288k23195365
You are doing fine. The total cost is $5000$ at the start and $36$ per year thereafter. The average cost starts very high and decreases as the $5000$ gets averaged over more years. It would be better to plot your curve over a range like $0-10$ or $0-30$, not up into the hundreds. You can write the average cost as $frac {5000}t+36$ and you can see that at $t=1$ it will be $5036$, at $t=2$ it will be $2518$ and so on.
answered Nov 16 at 3:15
Ross Millikan
288k23195365
answered Nov 16 at 3:15
Ross Millikan
288k23195365
answered Nov 16 at 3:15
Ross Millikan
288k23195365
answered Nov 16 at 3:15
Ross Millikan
288k23195365
288k23195365
add a comment |
add a comment |
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I mean ... it is a rational function, though? A rational function is a ratio of two polynomials. You have $$AC(t) = frac{4999.99 + 36t}{t}$$ Both the numerator and denominator are polynomials, so it is a rational function. The behavior of the graph even makes sense (if you take $t>0$ since "negative time" doesn't make sense) - the coefficient of $t$ in the numerator is small, so of course for very small $t$ (particularly $t<1$) you'll have a high value. I guess it might help to limit yourself to $tgeq 1$ since it wouldn't make sense to use intervals of years when it's not even 1 year.
– Eevee Trainer
Nov 16 at 3:03
Maybe I'm just not seeing what exactly the issue is. What kind of curve are you talking about?
– Eevee Trainer
Nov 16 at 3:03
@EeveeTrainer im confused as why the graph is decreasing, shouldn't the average cost of the product each year be greater than the starting price?
– xannax159
Nov 16 at 3:08
The thing is that, sure, you make a greater starting cost (which I assume to be the 4999.99 which I round to 5000 going forward), but you only pay 36 extra every successive year. So 1 year down the road, your total investment is only 5036. 2 years down the road, your total investment is 5072 - but over two years, that averages only 2536. After 3 years, the investment is 5108 - but over three years, that's an average of only about 1702. Notice the key point - that over time, your total investment increases, but your average investment gets smaller.
– Eevee Trainer
Nov 16 at 3:13
No, because you pay the initial 5000 only once.
– T. Bongers
Nov 16 at 3:13