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Find the tangent to this

$\displaystyle y={1 \over x+3}$

it's crossing the point $(-2,1)$

I have drawn the lines but I can't calculate it

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I formated your question in $\LaTeX$ please check that I didn't alter the meaning of what you asked. –  Git Gud Apr 1 '13 at 8:09
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You might find this useful: en.wikipedia.org/wiki/Derivation_(calculus) you can find lots of pictures of tangents here. –  kyticka Apr 1 '13 at 8:14
    
@Git Gud no you didn't. Thanks!! –  user1838781 Apr 1 '13 at 8:14
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2 Answers 2

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If you are expected to solve this graphically, you can extend your lines to the x and y axes and calculate the gradient from the two axis intercepts. The intercepts should be integers, so your graphs can be quite precise in this problem :)

If you draw the graphs of $y=\dfrac{1}{x+3}$ and drew a tangent line at the point $(−2,1)$, you would see that the tangent crosses the y-intercept at -1 and the x-intercept at -1. From this, you get the gradient of this tangent as -1 and the equation of the tangent becomes $y=−x−1$.

EDIT: Added stuff from my comment below, though it now gives the answer away...

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I don't think this answers the question. –  Git Gud Apr 1 '13 at 8:28
    
@Gid Gud You mentioned 'drawn the lines' which got me thinking that you have a graph drawn. If you draw the graphs of $y = \dfrac{1}{x+3}$ and drew a tangent line at the point $(-2,1)$, you would see that the tangent crosses the y-intercept at -1 and the x-intercept at -1. From this, you get the gradient of this tangent as -1 and the equation of the tangent becomes $y = -x - 1$. I think post did address your question. –  Jerry Apr 1 '13 at 8:37
    
Please note that I wasn't the one asking the question. The OP did mention he draw the lines, but I think that's just him telling the community what he tried. I don't think he's expected to solve it graphically. –  Git Gud Apr 1 '13 at 8:41
    
@GitGud And why not, may I ask? The tag calculus was not in the original post, and solving graphically are considered solutions as well. –  Jerry Apr 1 '13 at 8:44
    
As it is clear by my last comment, it is just my opinion. I can't explain why I think he's not supposed to solve it graphically, it's just the vibe I get from the question. Either way solving graphically is not solving at all (even if it happens to be what he's asked to do)! –  Git Gud Apr 1 '13 at 8:47
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Hint: The slope of the tangent line at the point $(-2, 1)$ is equal to $\dfrac{dy}{dx}(-2)$.

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Can you explaine that further? –  user1838781 Apr 3 '13 at 6:37
    
Since $\dfrac{dy}{dx} = \dfrac{-1}{(x + 3)^2}$, the slope of the tangent line should be equal to $\dfrac{dy}{dx}(-2) = -1$. So the required one is of the form $y = -x + c$ for some constant $c$. From the condition that the point $(-2, 1)$ on the line, $1 = -(-2) + c$ and hence $c = -1$. –  Taro Apr 3 '13 at 8:39
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