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For very large values of $x$, the graph of $f(x) = x^2 / (x + 1)$ behaves as that of a line $y = mx + b$. Find $m$ and $b$

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0

$begingroup$

A friend shared with me this problem and I'm not quite sure if I'm understanding it correctly. If the parabola behaves like a line, wouldn't it just be where they intersect? Graphing it we can see its $y=x-1$.

algebra-precalculus

share|cite|improve this question

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

$endgroup$

add a comment | 

0

$begingroup$

A friend shared with me this problem and I'm not quite sure if I'm understanding it correctly. If the parabola behaves like a line, wouldn't it just be where they intersect? Graphing it we can see its $y=x-1$.

algebra-precalculus

share|cite|improve this question

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

$endgroup$

add a comment | 

$begingroup$

A friend shared with me this problem and I'm not quite sure if I'm understanding it correctly. If the parabola behaves like a line, wouldn't it just be where they intersect? Graphing it we can see its $y=x-1$.

algebra-precalculus

share|cite|improve this question

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

$endgroup$

share|cite|improve this question

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

share|cite|improve this question

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

asked Mar 28 at 21:26

Tomás PalamásTomás Palamás

374211

2 Answers
2

active

oldest

votes

2

$begingroup$

We want to show $$lim_xtoinftyfracx^2x+1=lim_xtoinftyx-1$$

Observe that $$fracx^2x+1=fracx^2-1+1x+1=fracx^2-1x+1+frac1x+1=frac(x-1)(x+1)x+1+frac1x+1=x-1+frac1x+1$$

So, since $fracx^2x+1=x-1+frac1x+1$, we have that: $$lim_xtoinftyfracx^2x+1=lim_xtoinftyleft(x-1+frac1x+1right)=lim_xtoinftyleft(x-1right)+lim_xtoinftyfrac1x+1=lim_xtoinftyleft(x-1right)$$

which is what we wanted to show.

Edit:

The algebra way to think about this is to recognize the that $f(x)=fracx^2x+1$ is a rational function with $deg(textnumerator)=deg(textdenominator)+1$, hence we recognize that there exists an oblique asymptote to the rational function. This oblique asymptote is just the quotient of $$fracx^2x+1$$ which can be obtained by polynomial long division.

More information about oblique asymptotes can be found here.

share|cite|improve this answer

edited Mar 28 at 21:47

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible to do it without limits? This is my confusion because my friend isn't taking calculus, it's just an honors algebra class.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it back you answered my question, algebraically! The limits are unnecessary.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at my edit to think about this without limits (but limits are really lurking in disguise).
  $endgroup$
  – coreyman317
  Mar 28 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. When dividing, I forgot to divide $-x$ by $x+1$.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:52
  

  
  
  
  

add a comment | 

0

$begingroup$

$x^2=((x+1)-1)^2=$

$(x+1)^2-2(x+1)+1;$

$y=dfracx^2x+1= (x+1)-2 +dfrac1x+1;$

For large $x$ : $dfrac1x+1 rightarrow 0.$

Hence for large $x:$

$y=x-1;$

share|cite|improve this answer

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

$endgroup$

add a comment | 

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2

$begingroup$

We want to show $$lim_xtoinftyfracx^2x+1=lim_xtoinftyx-1$$

Observe that $$fracx^2x+1=fracx^2-1+1x+1=fracx^2-1x+1+frac1x+1=frac(x-1)(x+1)x+1+frac1x+1=x-1+frac1x+1$$

So, since $fracx^2x+1=x-1+frac1x+1$, we have that: $$lim_xtoinftyfracx^2x+1=lim_xtoinftyleft(x-1+frac1x+1right)=lim_xtoinftyleft(x-1right)+lim_xtoinftyfrac1x+1=lim_xtoinftyleft(x-1right)$$

which is what we wanted to show.

Edit:

The algebra way to think about this is to recognize the that $f(x)=fracx^2x+1$ is a rational function with $deg(textnumerator)=deg(textdenominator)+1$, hence we recognize that there exists an oblique asymptote to the rational function. This oblique asymptote is just the quotient of $$fracx^2x+1$$ which can be obtained by polynomial long division.

More information about oblique asymptotes can be found here.

share|cite|improve this answer

edited Mar 28 at 21:47

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible to do it without limits? This is my confusion because my friend isn't taking calculus, it's just an honors algebra class.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it back you answered my question, algebraically! The limits are unnecessary.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at my edit to think about this without limits (but limits are really lurking in disguise).
  $endgroup$
  – coreyman317
  Mar 28 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. When dividing, I forgot to divide $-x$ by $x+1$.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:52
  

  
  
  
  

add a comment | 

2

$begingroup$

We want to show $$lim_xtoinftyfracx^2x+1=lim_xtoinftyx-1$$

Observe that $$fracx^2x+1=fracx^2-1+1x+1=fracx^2-1x+1+frac1x+1=frac(x-1)(x+1)x+1+frac1x+1=x-1+frac1x+1$$

So, since $fracx^2x+1=x-1+frac1x+1$, we have that: $$lim_xtoinftyfracx^2x+1=lim_xtoinftyleft(x-1+frac1x+1right)=lim_xtoinftyleft(x-1right)+lim_xtoinftyfrac1x+1=lim_xtoinftyleft(x-1right)$$

which is what we wanted to show.

Edit:

The algebra way to think about this is to recognize the that $f(x)=fracx^2x+1$ is a rational function with $deg(textnumerator)=deg(textdenominator)+1$, hence we recognize that there exists an oblique asymptote to the rational function. This oblique asymptote is just the quotient of $$fracx^2x+1$$ which can be obtained by polynomial long division.

More information about oblique asymptotes can be found here.

share|cite|improve this answer

edited Mar 28 at 21:47

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible to do it without limits? This is my confusion because my friend isn't taking calculus, it's just an honors algebra class.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it back you answered my question, algebraically! The limits are unnecessary.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at my edit to think about this without limits (but limits are really lurking in disguise).
  $endgroup$
  – coreyman317
  Mar 28 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. When dividing, I forgot to divide $-x$ by $x+1$.
  $endgroup$
  – Tomás Palamás
  Mar 28 at 21:52
  

  
  
  
  

add a comment | 

$begingroup$

We want to show $$lim_xtoinftyfracx^2x+1=lim_xtoinftyx-1$$

Observe that $$fracx^2x+1=fracx^2-1+1x+1=fracx^2-1x+1+frac1x+1=frac(x-1)(x+1)x+1+frac1x+1=x-1+frac1x+1$$

So, since $fracx^2x+1=x-1+frac1x+1$, we have that: $$lim_xtoinftyfracx^2x+1=lim_xtoinftyleft(x-1+frac1x+1right)=lim_xtoinftyleft(x-1right)+lim_xtoinftyfrac1x+1=lim_xtoinftyleft(x-1right)$$

which is what we wanted to show.

Edit:

The algebra way to think about this is to recognize the that $f(x)=fracx^2x+1$ is a rational function with $deg(textnumerator)=deg(textdenominator)+1$, hence we recognize that there exists an oblique asymptote to the rational function. This oblique asymptote is just the quotient of $$fracx^2x+1$$ which can be obtained by polynomial long division.

More information about oblique asymptotes can be found here.

share|cite|improve this answer

edited Mar 28 at 21:47

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

$endgroup$

share|cite|improve this answer

edited Mar 28 at 21:47

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

answered Mar 28 at 21:35

coreyman317coreyman317

1,256522

0

$begingroup$

$x^2=((x+1)-1)^2=$

$(x+1)^2-2(x+1)+1;$

$y=dfracx^2x+1= (x+1)-2 +dfrac1x+1;$

For large $x$ : $dfrac1x+1 rightarrow 0.$

Hence for large $x:$

$y=x-1;$

share|cite|improve this answer

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

$endgroup$

add a comment | 

0

$begingroup$

$x^2=((x+1)-1)^2=$

$(x+1)^2-2(x+1)+1;$

$y=dfracx^2x+1= (x+1)-2 +dfrac1x+1;$

For large $x$ : $dfrac1x+1 rightarrow 0.$

Hence for large $x:$

$y=x-1;$

share|cite|improve this answer

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

$endgroup$

add a comment | 

$begingroup$

$x^2=((x+1)-1)^2=$

$(x+1)^2-2(x+1)+1;$

$y=dfracx^2x+1= (x+1)-2 +dfrac1x+1;$

For large $x$ : $dfrac1x+1 rightarrow 0.$

Hence for large $x:$

$y=x-1;$

share|cite|improve this answer

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

$endgroup$

share|cite|improve this answer

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822

answered Mar 28 at 22:15

Peter SzilasPeter Szilas

11.7k2822