Finding the Asymptote / Root of a reciprocal function
$begingroup$
$$y = frac{3}{8x - 3} $$
The y-intercept is $-1$ and the vertical asymptote is $x = frac{3}{8}$ but what would be the horizontal asymptote and the x-intercept in this case?
I am asking this as the numerator is not a linear function (it is just a constant 3 so I do not know what the horizontal asymptote or root would be). I follow the procedure below:
$$y = frac{ax + b}{cx + d}$$
root at $x = frac{-b}{a}$
intercept at $y = frac{b}{d} $
vertical asymptote at $x = frac{-d}{c} $
horizontal asymptote at $y = frac{a}{c}$
When finding the root, you get $0 = frac{3}{8x - 3}$ and then $0 = 3$ which is not true, therefore this must mean the curve does not cut the x-axis?
The horizontal asymptote (using what I posted above) would be $ y = frac{a}{c}$ which is $frac{0}{8}$ hence the horizontal asymptote is $ y = 0$?
functions graphing-functions
asked Apr 15 '14 at 12:35
NoahNoah
3914
$endgroup$
add a comment |
$begingroup$
$$y = frac{3}{8x - 3} $$
The y-intercept is $-1$ and the vertical asymptote is $x = frac{3}{8}$ but what would be the horizontal asymptote and the x-intercept in this case?
I am asking this as the numerator is not a linear function (it is just a constant 3 so I do not know what the horizontal asymptote or root would be). I follow the procedure below:
$$y = frac{ax + b}{cx + d}$$
root at $x = frac{-b}{a}$
intercept at $y = frac{b}{d} $
vertical asymptote at $x = frac{-d}{c} $
horizontal asymptote at $y = frac{a}{c}$
When finding the root, you get $0 = frac{3}{8x - 3}$ and then $0 = 3$ which is not true, therefore this must mean the curve does not cut the x-axis?
The horizontal asymptote (using what I posted above) would be $ y = frac{a}{c}$ which is $frac{0}{8}$ hence the horizontal asymptote is $ y = 0$?
functions graphing-functions
asked Apr 15 '14 at 12:35
NoahNoah
3914
$endgroup$
add a comment |
$begingroup$
$$y = frac{3}{8x - 3} $$
The y-intercept is $-1$ and the vertical asymptote is $x = frac{3}{8}$ but what would be the horizontal asymptote and the x-intercept in this case?
I am asking this as the numerator is not a linear function (it is just a constant 3 so I do not know what the horizontal asymptote or root would be). I follow the procedure below:
$$y = frac{ax + b}{cx + d}$$
root at $x = frac{-b}{a}$
intercept at $y = frac{b}{d} $
vertical asymptote at $x = frac{-d}{c} $
horizontal asymptote at $y = frac{a}{c}$
When finding the root, you get $0 = frac{3}{8x - 3}$ and then $0 = 3$ which is not true, therefore this must mean the curve does not cut the x-axis?
The horizontal asymptote (using what I posted above) would be $ y = frac{a}{c}$ which is $frac{0}{8}$ hence the horizontal asymptote is $ y = 0$?
functions graphing-functions
asked Apr 15 '14 at 12:35
NoahNoah
3914
$endgroup$
$$y = frac{3}{8x - 3} $$
The y-intercept is $-1$ and the vertical asymptote is $x = frac{3}{8}$ but what would be the horizontal asymptote and the x-intercept in this case?
I am asking this as the numerator is not a linear function (it is just a constant 3 so I do not know what the horizontal asymptote or root would be). I follow the procedure below:
$$y = frac{ax + b}{cx + d}$$
root at $x = frac{-b}{a}$
intercept at $y = frac{b}{d} $
vertical asymptote at $x = frac{-d}{c} $
horizontal asymptote at $y = frac{a}{c}$
When finding the root, you get $0 = frac{3}{8x - 3}$ and then $0 = 3$ which is not true, therefore this must mean the curve does not cut the x-axis?
The horizontal asymptote (using what I posted above) would be $ y = frac{a}{c}$ which is $frac{0}{8}$ hence the horizontal asymptote is $ y = 0$?
functions graphing-functions
functions graphing-functions
asked Apr 15 '14 at 12:35
NoahNoah
3914
asked Apr 15 '14 at 12:35
NoahNoah
3914
edited Apr 15 '14 at 13:18
Noah
asked Apr 15 '14 at 12:35
NoahNoah
3914
asked Apr 15 '14 at 12:35
NoahNoah
3914
asked Apr 15 '14 at 12:35
NoahNoah
3914
3914
add a comment |
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2 Answers
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$begingroup$
For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $infty$), or when it gets smaller and smaller (towards $-infty$).
Note that $frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.
Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=frac{0}{c}=0$, just as it should be.
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
$endgroup$
add a comment |
$begingroup$
When the degree of the denominator is greater than the degree of the numerator, limits at $infty$ and at $-infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
$endgroup$
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $infty$), or when it gets smaller and smaller (towards $-infty$).
Note that $frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.
Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=frac{0}{c}=0$, just as it should be.
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
$endgroup$
add a comment |
$begingroup$
For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $infty$), or when it gets smaller and smaller (towards $-infty$).
Note that $frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.
Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=frac{0}{c}=0$, just as it should be.
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
$endgroup$
add a comment |
$begingroup$
For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $infty$), or when it gets smaller and smaller (towards $-infty$).
Note that $frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.
Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=frac{0}{c}=0$, just as it should be.
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
$endgroup$
For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $infty$), or when it gets smaller and smaller (towards $-infty$).
Note that $frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.
Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=frac{0}{c}=0$, just as it should be.
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
answered Apr 15 '14 at 14:06
naslundxnaslundx
8,00452941
8,00452941
add a comment |
add a comment |
$begingroup$
When the degree of the denominator is greater than the degree of the numerator, limits at $infty$ and at $-infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
$endgroup$
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
add a comment |
$begingroup$
When the degree of the denominator is greater than the degree of the numerator, limits at $infty$ and at $-infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
$endgroup$
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
add a comment |
$begingroup$
When the degree of the denominator is greater than the degree of the numerator, limits at $infty$ and at $-infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
$endgroup$
When the degree of the denominator is greater than the degree of the numerator, limits at $infty$ and at $-infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
answered Apr 15 '14 at 12:42
ajotatxeajotatxe
54.2k24190
54.2k24190
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
add a comment |
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
$begingroup$
I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it).
$endgroup$
– Noah
Apr 15 '14 at 13:19
add a comment |
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