Why is the y-intercept of this calculus problem given like that in the solution? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Limit with square and cube root difference $lim_n to infty left(sqrtn^2 + n - sqrt[3]n^3 + n^2right)$How do I get the equation of the line given this slope and graph?calculus where m is the slope and b is the y interceptFinding the tangent line to a curve at a given point? Stumped by simple problem.Finding Equation of tangent lineCalculus problem of finding the equation of a line.how do I find the tangent line of a function that passes a given pointTangent line triangle area proof request.Equation for the line of a derivativeGiven two points that are joined by a line that is a tangent to a curve, find the missing constant in the equation for the curve
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Why is the y-intercept of this calculus problem given like that in the solution?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Limit with square and cube root difference $lim_n to infty left(sqrtn^2 + n - sqrt[3]n^3 + n^2right)$How do I get the equation of the line given this slope and graph?calculus where m is the slope and b is the y interceptFinding the tangent line to a curve at a given point? Stumped by simple problem.Finding Equation of tangent lineCalculus problem of finding the equation of a line.how do I find the tangent line of a function that passes a given pointTangent line triangle area proof request.Equation for the line of a derivativeGiven two points that are joined by a line that is a tangent to a curve, find the missing constant in the equation for the curve
$begingroup$
Given the following problem:
Let $f$ be the real-valued function defined by $f(x) = sqrt1+6x$.
Determine the slope of the line tangent to the graph of $f$ at $x=4$.
Determine the y-intercept of the line tangent to the graph of $f$ at $x=4$.
For the life of my I can't work out why the solution to the $y$ intercept is given as $frac135$.
I've correctly solved $f'$ using the definition of a derivative formula (I got $frac3sqrt1+6x$) and do get the correct answer for the slope of the line tangent to $x=4$ (which is $frac35$) but the $y$ intercept for that line is definitely $3$ and not $frac135$. Where am I going wrong? This is the solution that is given by the authors:
calculus chain-rule tangent-line
asked Mar 31 at 14:30
NobilisNobilis
1054
$endgroup$
add a comment |
$begingroup$
Given the following problem:
Let $f$ be the real-valued function defined by $f(x) = sqrt1+6x$.
Determine the slope of the line tangent to the graph of $f$ at $x=4$.
Determine the y-intercept of the line tangent to the graph of $f$ at $x=4$.
For the life of my I can't work out why the solution to the $y$ intercept is given as $frac135$.
I've correctly solved $f'$ using the definition of a derivative formula (I got $frac3sqrt1+6x$) and do get the correct answer for the slope of the line tangent to $x=4$ (which is $frac35$) but the $y$ intercept for that line is definitely $3$ and not $frac135$. Where am I going wrong? This is the solution that is given by the authors:
calculus chain-rule tangent-line
asked Mar 31 at 14:30
NobilisNobilis
1054
$endgroup$
$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29
add a comment |
$begingroup$
Given the following problem:
Let $f$ be the real-valued function defined by $f(x) = sqrt1+6x$.
Determine the slope of the line tangent to the graph of $f$ at $x=4$.
Determine the y-intercept of the line tangent to the graph of $f$ at $x=4$.
For the life of my I can't work out why the solution to the $y$ intercept is given as $frac135$.
I've correctly solved $f'$ using the definition of a derivative formula (I got $frac3sqrt1+6x$) and do get the correct answer for the slope of the line tangent to $x=4$ (which is $frac35$) but the $y$ intercept for that line is definitely $3$ and not $frac135$. Where am I going wrong? This is the solution that is given by the authors:
calculus chain-rule tangent-line
asked Mar 31 at 14:30
NobilisNobilis
1054
$endgroup$
Given the following problem:
Let $f$ be the real-valued function defined by $f(x) = sqrt1+6x$.
Determine the slope of the line tangent to the graph of $f$ at $x=4$.
Determine the y-intercept of the line tangent to the graph of $f$ at $x=4$.
For the life of my I can't work out why the solution to the $y$ intercept is given as $frac135$.
I've correctly solved $f'$ using the definition of a derivative formula (I got $frac3sqrt1+6x$) and do get the correct answer for the slope of the line tangent to $x=4$ (which is $frac35$) but the $y$ intercept for that line is definitely $3$ and not $frac135$. Where am I going wrong? This is the solution that is given by the authors:
calculus chain-rule tangent-line
calculus chain-rule tangent-line
asked Mar 31 at 14:30
NobilisNobilis
1054
asked Mar 31 at 14:30
NobilisNobilis
1054
asked Mar 31 at 14:30
NobilisNobilis
1054
asked Mar 31 at 14:30
NobilisNobilis
1054
asked Mar 31 at 14:30
NobilisNobilis
1054
1054
$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29
add a comment |
$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29
$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29
$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Say the tangent line is $y=mx+b$ with $m=frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
$endgroup$
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Say the tangent line is $y=mx+b$ with $m=frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
$endgroup$
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
add a comment |
$begingroup$
Say the tangent line is $y=mx+b$ with $m=frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
$endgroup$
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
add a comment |
$begingroup$
Say the tangent line is $y=mx+b$ with $m=frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
$endgroup$
Say the tangent line is $y=mx+b$ with $m=frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
answered Mar 31 at 14:34
J. W. TannerJ. W. Tanner
4,7721420
4,7721420
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
add a comment |
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
$begingroup$
Ah, but isn't the y-intercept defined at $x=0$?
$endgroup$
– Nobilis
Mar 31 at 14:37
1
1
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
$begingroup$
Yes, the solution is $b=frac135$ so the equation for tangent line is $y=frac35x+frac135$, and, when $x=0$, $y=frac135$
$endgroup$
– J. W. Tanner
Mar 31 at 14:39
1
1
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
I gave the slope-intercept form of the line ($b$ is the y-intercept)
$endgroup$
– J. W. Tanner
Mar 31 at 14:42
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
$begingroup$
Right, I get it, $y=3$ is the y-intercept of the derivative, not the tangent line (I plotted the two of them and it now makes sense).
$endgroup$
– Nobilis
Mar 31 at 14:45
add a comment |
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$begingroup$
When you ask why your answer differs from the expected one, it’s very helpful to show your work instead of simply declaring your result. That way, you’re not making people who want to help you guess where you might have gone wrong.
$endgroup$
– amd
Mar 31 at 18:29