# Calculus / tangent line

Find the exact value of $c$ in the figure shown below, where the line $l$ tangent to the graph of $y = 2^x$ at $(0, 1)$ intersects the $x$-axis.

looking at the graph, find the exact slope of the tangent line at $(0, 1)$, the equation of the tangent line and the exact value of $c$.

I started with finding the derivative of the equation, so $y= 2^x$

$y'=\ln(2) 2^x$

$f'(0)= \ln(2)$, is that right for the slope ?

tangent equation = $y=mx+b$

$y= \ln(2)x+1$

I feel that I am far away from the right answer. could anyone help me get through it.

• The slope is right. To get $c$ (the $x$-intercept) set $y=0$ in the equation of the tangent line, and solve for $x$. – André Nicolas Jun 19 '14 at 0:22

I can't read what you have, but

$$y' = 2^x \ln 2$$

and $y'(0) = \ln2$

so $$l - 1 = \ln 2 x \implies l = (\ln 2)x+1$$

• Thanks, so it seems that mine is right – John Jun 19 '14 at 0:24

I think your equation of the tangent line is correct. To find c just plug in $y=0$ and get $x$. Then the $x$ you get is the $c$ you want to find.

• Yeah! I thought of it, but i wasn't sure about my equation!thanks – John Jun 19 '14 at 0:24

First of all, differentiate $2^x$, as follows: $$\color{green}{\frac{d}{dx}[2^x]}=\frac{d}{dx}[e^{x\ln(2)}]=\ln(2)e^{x\ln(2)}\color{green}{=2^x\ln(2)}.$$

To find the gradient of the tangent line at $x=0$, $m$, evaluate $f'(0)$.

Then, to find the equation of the tangent line, use the formula $$y-y_1=m[x-x_1]$$ where $(x_1,y_1)=(0,1)$.

Once you've got the equation of the tangent line, in the form $y=ax+b$ (where $a$ and $b$ are to be determined), to find the $x-$intercept, set $y=0$, so $ax+b=0 \iff x=-\frac{b}{a}$.

Your job, now, is to find $a$ and $b$ (i.e. the equation of the tangent line).