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Given the graph of $y=-e^{-x}$ and that there is a tangent to the graph that crosses the x axis at $(-4,0)$ determine the slope of that line.

So this seems like a simple question but I don't know why I'm racking my brain for this. I know I need to know the $(x,y)$ cordinate of where the line touches the graph, but I have no idea how to find it.

I do know that once I find the coordinates, I can simply differentiate $y=-e^{-x}$ and find the slope of the tangent line.

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Well $y$ is never $0$, it just gets closer and closer to $0$. You can take a look at the graph, it never touches the $x$ axis: wolframalpha.com/input/?i=y+%3D+-e%5E-x –  Dane Bouchie Jul 31 at 19:47
    
@DaneBouchie how would that help –  Varun Iyer Jul 31 at 19:47
    
That doesn't help because (-4,0) is the x-intersection of the $tangent$ line –  Panthy Jul 31 at 19:48
    
I think there actually might be some ambiguity in whether the "that" in the problem statement is "the graph" or "a tangent." I think the intended reading is the tangent. –  Jason Knapp Jul 31 at 19:49
    
Ah, so to answer your question... Just use the point-slope form, and use the derivative as the slope: $y-y_0 = m(x-x_0)$, then solve for $y$. –  Dane Bouchie Jul 31 at 19:50

4 Answers 4

up vote 1 down vote accepted

Let's call the slope of our line $m$.

We know that for some value of x,

$$f'(x) = e^{-x} = m$$

Expressing our line in point-slope form:

$$y-0 = m(x+4)$$

Substituting:

$$y= m(x+4)$$

$$y = e^{-x}(x+4)$$

Also, note that the point of tangency can be expressed as $(x, -e^{-x})$

Using this:

$$-e^{-x} = e^{-x}(x+4)$$

Solving for $x$, we get that $$x = -5$$

Therefore, the slope of our line is:

$$f'(-5) = e^5$$

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A point on the graph would look like $(x, -e^{-x})$. So the slope of the line passing through $(-4,0)$ and $(x, -e^{-x})$ would be

$$m=\frac{-e^{-x}}{x+4}$$

And we want this line to be tangent to $-e^{-x}$ at $(x, -e^{-x})$ so we want $m = e^{-x}$. So we need to solve this:

$$e^{-x} =\frac{-e^{-x}}{x+4} $$

Hence $x+4 = -1$ or $x = -5$. And so the slope would be $e^{5}$.

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Its $x+4$ not $x-4$ –  Varun Iyer Jul 31 at 19:52
    
Thanks, I rewrote the -4 as a 4 and the typo carried through. I fixed it now. –  Mastrel Jul 31 at 19:54
    
no problem. Always happy to help :) –  Varun Iyer Jul 31 at 19:54

Let the point of tangency be $(a,e^{-a})$. Then the tangent line has slope $-e^{-a}$.

But the tangent line goes through $(a,e^{-a})$ and $(-4,0)$, so it has slope $\dfrac{e^{-a}-0}{a-(-4)}$.

Set the two expressions for slope equal to each other, and solve for $a$.

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It sounds like you are comfortable with finding a tangent line given a point on the graph, so here is an idea:

The tangent line at $(a, f(a))$ will have the form (point-slope or one of those names) $$y - f(a) = e^{-a}(x-a)$$ Can you find a value of $a$ so that this tangent line has the required intercept?

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I originally thought of doing that like this: $y-0=\left(\frac{d}{dx}-e^{-x}\right)(x+4)$ –  Panthy Jul 31 at 19:51
    
Two comments. 1: to be specific, on the right it is $\frac{d}{dx}(-e^{-x})$, the $(x+4)$ is excluded from the derivative of course. 2: On the left, $f(a)$ is still just $f(a)$, but $y$ is $0$ because the point on the tangent line is $(-4,0)$. –  Jason Knapp Jul 31 at 19:52
    
Gosh sorry I wrote so hastily. It is important to keep the $a$ and the $x$ straight. $a$ is the $x-$value where the tangent line is.. tangent. $x$ is the value along the tangent line. Therefore it is important as in my answer that the "slope" is $e^{-a}$ and the linear part is $(x+4)$ Does this help at all? –  Jason Knapp Jul 31 at 19:56

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