Find the equation of the tangent to the curve with exponential function

Question

The question is as follows: Find the equation of the tangent to the curve $y = xe^{2x}$ at the point $(\frac{1}{2}, \frac{e}{2})$.

Now I figured out that $\frac{dy}{dx} = e^{2x}(2x+1)$, and that when I plug in $x=1/2$ then I get that the slope = $2e$.

So at this point I have the original curve's equation, the equation of its differential, the fact that the slope of the tangent at the given point is $2e$ and that this tangent also passes through the point $(\frac{1}{2}, \frac{e}{2})$. But I can't seem to arrive at the equation of this tangent.

The answer is

\begin{equation} y = 2ex - \frac{e}{2} \end{equation}

but how they got there, I don't know. I've checked other find the equation of a tangent line to a curve questions, but still haven't figured my way to that answer. It seems there's something wrong with my assumption that the equation of the tangent line is of the form $y=mx+c$. But how do I know which form it should take?

Edit

Sorry - I'd written the target answer above wrong. I edited it to correct it.

Answer 1

The equation of a line of slope $m$ passing through a point $(x_0,y_0)$ is

$$y-y_0 = m (x - x_0)$$

Here, $m=2 e$, $x_0 = \frac{1}{2}$, and $y_0 = \frac{e}{2}$. Plug away.