find two equations for the tangent lines to the curve
$begingroup$
So I have one answer but I can't figure out how to get the other answer so I was seeing if someone could help me out here.
Find equations of the tangent lines to the curve $displaystyle y=frac{x-1}{x+1}$ that are parallel to the line $x-2y=5$. Then it wants you to find $y=$ _______ (smaller $y$-intercept) and $y=$ ________ (larger $y$-intercept). I have found the larger $y$-intercept which is $displaystyle y=frac{1}{2}x+frac{7}{2}$, but I'm not sure how I am supposed to find the smaller $y$-intercept.
calculus algebra-precalculus
asked Oct 5 '14 at 20:44
rickrick
77210
$endgroup$
add a comment |
$begingroup$
So I have one answer but I can't figure out how to get the other answer so I was seeing if someone could help me out here.
Find equations of the tangent lines to the curve $displaystyle y=frac{x-1}{x+1}$ that are parallel to the line $x-2y=5$. Then it wants you to find $y=$ _______ (smaller $y$-intercept) and $y=$ ________ (larger $y$-intercept). I have found the larger $y$-intercept which is $displaystyle y=frac{1}{2}x+frac{7}{2}$, but I'm not sure how I am supposed to find the smaller $y$-intercept.
calculus algebra-precalculus
asked Oct 5 '14 at 20:44
rickrick
77210
$endgroup$
$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53
add a comment |
$begingroup$
So I have one answer but I can't figure out how to get the other answer so I was seeing if someone could help me out here.
Find equations of the tangent lines to the curve $displaystyle y=frac{x-1}{x+1}$ that are parallel to the line $x-2y=5$. Then it wants you to find $y=$ _______ (smaller $y$-intercept) and $y=$ ________ (larger $y$-intercept). I have found the larger $y$-intercept which is $displaystyle y=frac{1}{2}x+frac{7}{2}$, but I'm not sure how I am supposed to find the smaller $y$-intercept.
calculus algebra-precalculus
asked Oct 5 '14 at 20:44
rickrick
77210
$endgroup$
So I have one answer but I can't figure out how to get the other answer so I was seeing if someone could help me out here.
Find equations of the tangent lines to the curve $displaystyle y=frac{x-1}{x+1}$ that are parallel to the line $x-2y=5$. Then it wants you to find $y=$ _______ (smaller $y$-intercept) and $y=$ ________ (larger $y$-intercept). I have found the larger $y$-intercept which is $displaystyle y=frac{1}{2}x+frac{7}{2}$, but I'm not sure how I am supposed to find the smaller $y$-intercept.
calculus algebra-precalculus
calculus algebra-precalculus
asked Oct 5 '14 at 20:44
rickrick
77210
asked Oct 5 '14 at 20:44
rickrick
77210
edited Oct 5 '14 at 22:43
rick
asked Oct 5 '14 at 20:44
rickrick
77210
asked Oct 5 '14 at 20:44
rickrick
77210
asked Oct 5 '14 at 20:44
rickrick
77210
77210
$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53
add a comment |
$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53
$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
$endgroup$
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
add a comment |
Your Answer
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1 Answer
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votes
1 Answer
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$begingroup$
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
$endgroup$
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
add a comment |
$begingroup$
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
$endgroup$
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
add a comment |
$begingroup$
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
$endgroup$
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
answered Oct 5 '14 at 22:58
Rory DaultonRory Daulton
29.6k63355
29.6k63355
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
add a comment |
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
yes I did but my other x was 1. So when you plug that in it is 0/2. Or is it not 1?
$endgroup$
– rick
Oct 5 '14 at 23:03
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
Yes, the other $x$ is $1$. Plug that into the original equation and you get $y=0$. Use the resulting point $(1,0)$ in the point-slope form of a line to get the other desired tangent line.
$endgroup$
– Rory Daulton
Oct 5 '14 at 23:08
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
$begingroup$
ok thank you I got it now.
$endgroup$
– rick
Oct 5 '14 at 23:18
add a comment |
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$begingroup$
It is not clear what you are asking. A $y$-intercept is not an equation, it is a real number: the value of $y$ for a point on the curve on the $y$-axis.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:28
$begingroup$
@RoryDaulton I added the picture that's exactly how the question is displayed.
$endgroup$
– rick
Oct 5 '14 at 22:44
$begingroup$
@RoryDaulton I know I need to use the quotient rule and finding the derivative is the slope. I just don't know how you are supposed to get two equations from it.
$endgroup$
– rick
Oct 5 '14 at 22:45
$begingroup$
I see. You do not want the $y$-intercept, you want the equation that has the lower $y$-intercept. See my answer for more comments.
$endgroup$
– Rory Daulton
Oct 5 '14 at 22:53