# Thread: I need some help with a tangent line problem

1. ## I need some help with a tangent line problem

Hi.

I need some HW help again.

Let f(x) = 12x + 15 - 13e^x . Then the equation of the tangent line to the graph of f(x) at the point (0, 2) is given by y=mx+b where:

m = ?

For f'(x), I get: 12 - 13e^x

which I would think is the slope, but I can't get it to work out.

Thanks

2. Since you are already given the point $\displaystyle (0,2)$ on the curve $\displaystyle f(x)$, all you really need is to find $\displaystyle m$, and then plug in the point $\displaystyle (0,2)$ and $\displaystyle m$ into $\displaystyle y = mx + b$ to solve for $\displaystyle b$. Notice that $\displaystyle m = f'(0)$, and $\displaystyle f'(0) = 12 - 13(e^0) = 12 - 13 = -1$. Also note that $\displaystyle y = -x + b$ since $\displaystyle m = -1$. Plugging in $\displaystyle 0$ for $\displaystyle x$ and 2 for $\displaystyle y$ we see that $\displaystyle 2 = -(0) + b$, which means that $\displaystyle b = 2$. This in turn means that $\displaystyle y = -1(x) + 2$, and you are done!

3. Wow. was that really all I had to do? plug in a 0 for the x? It all makes sense now but I was working on it forever and for some reason that never occurred to me.

Anyway, thank you so much.