Calculating tangent line at $t=2$ 
Calculate the tangent line at $t=2$ of the curve
  $$x(t)=t^2+2t+4,\ y(t)=1+te^t$$

How would you go about determining this? So far I have
$$x'(t)=2t + 2\\
y'(t) = te^t+e^t$$
Where do I go from here to get the tangent line?

The following expression is how you can find the slope of a parametric curve:
$$
\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)},\ \frac{dx}{dt}\ne0.
$$
When you have found the slope of the curve at $t=2$, use the equation of a straight line to form the actual line:
$$
y-y_0=m(x-x_0).
$$
Use your original parametric equation to find the point $(x_0,y_0)$ at $t=2$.
