1. Tangent line

I have a calc test tomorrow and I'm stuck on this practice problem. Please help. He gave us the answer which is y=2x+1, but I don't know how to do it. Please help. I have an idea of how he got it, but the e threw me off.

Find the indicated tangent line.

Find the tangent line to the graph of f(x) = e^2x at the point (0, 1).

2. Originally Posted by aphan19
I have a calc test tomorrow and I'm stuck on this practice problem. Please help. He gave us the answer which is y=2x+1, but I don't know how to do it. Please help. I have an idea of how he got it, but the e threw me off.

Find the indicated tangent line.

Find the tangent line to the graph of f(x) = e^2x at the point (0, 1).

Use the formula for a tangent line

$\displaystyle y-f(x_0)=f'(x_0)(x-x_0)$

Using $\displaystyle x_0=0$

$\displaystyle f(x_0)=1$

and $\displaystyle f'(x_0)=f'(0)$
were $\displaystyle f'(x)=2e^{2x}$

3. Originally Posted by aphan19
I have a calc test tomorrow and I'm stuck on this practice problem. Please help. He gave us the answer which is y=2x+1, but I don't know how to do it. Please help. I have an idea of how he got it, but the e threw me off.

Find the indicated tangent line.

Find the tangent line to the graph of f(x) = e^2x at the point (0, 1).

To find the tangent line, we first need to know the slope. We can find out the slope by finding f'(x) at x=0.

$\displaystyle \frac{dy}{dx}=2e^{2x}$. At $\displaystyle x=0$, we get $\displaystyle \frac{dy}{dx}=2$.

Now that we have the slope, now find the equation of the tangent line:

$\displaystyle y-(1)=(2)(x-(0))\rightarrow y=2x+1$.