Question about tangent and slope 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.
 A: 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 $e^{5}$.
A: 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 $a$.
A: 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?
A: 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:
$$f'(-5) = e^5$$
