Finding equation of a curve given the gradient of the tangent to the curve
$begingroup$
If the gradient of the tangent to a curve is given by $2x+1$ and the curve passes through the point $(-3,0)$, find the equation of the curve.
Shouldn't it be the equation of the tangent to the curve since the gradient in a linear equation of the form $ax+b$ is represented by $a$?
calculus integration
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
$endgroup$
add a comment |
$begingroup$
If the gradient of the tangent to a curve is given by $2x+1$ and the curve passes through the point $(-3,0)$, find the equation of the curve.
Shouldn't it be the equation of the tangent to the curve since the gradient in a linear equation of the form $ax+b$ is represented by $a$?
calculus integration
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
$endgroup$
5
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49
add a comment |
$begingroup$
If the gradient of the tangent to a curve is given by $2x+1$ and the curve passes through the point $(-3,0)$, find the equation of the curve.
Shouldn't it be the equation of the tangent to the curve since the gradient in a linear equation of the form $ax+b$ is represented by $a$?
calculus integration
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
$endgroup$
If the gradient of the tangent to a curve is given by $2x+1$ and the curve passes through the point $(-3,0)$, find the equation of the curve.
Shouldn't it be the equation of the tangent to the curve since the gradient in a linear equation of the form $ax+b$ is represented by $a$?
calculus integration
calculus integration
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
edited Oct 22 '17 at 8:08
galaxymcpvp
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
asked Oct 22 '17 at 8:01
galaxymcpvpgalaxymcpvp
444
444
5
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49
add a comment |
5
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49
5
5
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You're given: $frac{dy}{dx}=2x+1$
By simple integration you obtain:
$$y=x^2+x-6$$
Now, you say that:
Got the function [...], but when I graphed both functions, I got two intersections
This is obviously a misunderstood claim. The expression $2x+1$ gives you the slope of the tangent at a point $x$ of the graph, it does NOT give you the equation of the tangent at that point.
For a comparison, at point $P(-3,0)$, the slope of the tangent is $m=2cdot(-3)+1=-5$ and so, the equation of the tangent is $y-0=-5cdot(x-(-3))$. You can verify it from graph as well:
Also, note that different points of the curve will have different slopes and correspondingly different equations for tangents.
The line $y=2x+1$, thus, is in no way a "tangent" to the graph of $y$.
Hope your query is clarified!
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
$endgroup$
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You're given: $frac{dy}{dx}=2x+1$
By simple integration you obtain:
$$y=x^2+x-6$$
Now, you say that:
Got the function [...], but when I graphed both functions, I got two intersections
This is obviously a misunderstood claim. The expression $2x+1$ gives you the slope of the tangent at a point $x$ of the graph, it does NOT give you the equation of the tangent at that point.
For a comparison, at point $P(-3,0)$, the slope of the tangent is $m=2cdot(-3)+1=-5$ and so, the equation of the tangent is $y-0=-5cdot(x-(-3))$. You can verify it from graph as well:
Also, note that different points of the curve will have different slopes and correspondingly different equations for tangents.
The line $y=2x+1$, thus, is in no way a "tangent" to the graph of $y$.
Hope your query is clarified!
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
$endgroup$
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
add a comment |
$begingroup$
You're given: $frac{dy}{dx}=2x+1$
By simple integration you obtain:
$$y=x^2+x-6$$
Now, you say that:
Got the function [...], but when I graphed both functions, I got two intersections
This is obviously a misunderstood claim. The expression $2x+1$ gives you the slope of the tangent at a point $x$ of the graph, it does NOT give you the equation of the tangent at that point.
For a comparison, at point $P(-3,0)$, the slope of the tangent is $m=2cdot(-3)+1=-5$ and so, the equation of the tangent is $y-0=-5cdot(x-(-3))$. You can verify it from graph as well:
Also, note that different points of the curve will have different slopes and correspondingly different equations for tangents.
The line $y=2x+1$, thus, is in no way a "tangent" to the graph of $y$.
Hope your query is clarified!
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
$endgroup$
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
add a comment |
$begingroup$
You're given: $frac{dy}{dx}=2x+1$
By simple integration you obtain:
$$y=x^2+x-6$$
Now, you say that:
Got the function [...], but when I graphed both functions, I got two intersections
This is obviously a misunderstood claim. The expression $2x+1$ gives you the slope of the tangent at a point $x$ of the graph, it does NOT give you the equation of the tangent at that point.
For a comparison, at point $P(-3,0)$, the slope of the tangent is $m=2cdot(-3)+1=-5$ and so, the equation of the tangent is $y-0=-5cdot(x-(-3))$. You can verify it from graph as well:
Also, note that different points of the curve will have different slopes and correspondingly different equations for tangents.
The line $y=2x+1$, thus, is in no way a "tangent" to the graph of $y$.
Hope your query is clarified!
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
$endgroup$
You're given: $frac{dy}{dx}=2x+1$
By simple integration you obtain:
$$y=x^2+x-6$$
Now, you say that:
Got the function [...], but when I graphed both functions, I got two intersections
This is obviously a misunderstood claim. The expression $2x+1$ gives you the slope of the tangent at a point $x$ of the graph, it does NOT give you the equation of the tangent at that point.
For a comparison, at point $P(-3,0)$, the slope of the tangent is $m=2cdot(-3)+1=-5$ and so, the equation of the tangent is $y-0=-5cdot(x-(-3))$. You can verify it from graph as well:
Also, note that different points of the curve will have different slopes and correspondingly different equations for tangents.
The line $y=2x+1$, thus, is in no way a "tangent" to the graph of $y$.
Hope your query is clarified!
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
answered Oct 22 '17 at 9:11
Gaurang TandonGaurang Tandon
3,50432150
3,50432150
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
add a comment |
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
Ahh! I see. Thanks for clarifying!
$endgroup$
– galaxymcpvp
Oct 22 '17 at 11:03
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
$begingroup$
@galaxymcpvp Please accept the answer if you find it useful :)
$endgroup$
– Gaurang Tandon
Oct 22 '17 at 11:05
add a comment |
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5
$begingroup$
I think the question means that you have a function $f:mathbb{R}rightarrowmathbb{R}$ with a derivative $f'(x) = 2x+1$ for all $x in mathbb{R}$, and you also know $f(-3)=0$. So then find the function.
$endgroup$
– Michael
Oct 22 '17 at 8:04
$begingroup$
Got the function $y=x^2+x-6$, but when I graphed both functions, I got two intersections. As seen here: desmos.com/calculator/x6vttaawgl
$endgroup$
– galaxymcpvp
Oct 22 '17 at 8:49