Answer
The equation of the tangent line to the curve at $(0,2)$ is $$(l):y=3x+2$$
Work Step by Step
$$y=2e^x+x$$
1) First, find the derivative of y $$y'=2\frac{d}{dx}(e^x)+\frac{d}{dx}x$$
$$y'=2e^x+1$$
2) Find the slope of the tangent line to the curve at the point $(0,2)$, or in fact, $y′(0)$: $$y′(0)=2(e^0)+1=2\times1+1=3$$
3) The tangent line to the curve at the point $(0,2)$ would have the following equation: $$(l):y=y′(0)x+b$$
$$(l):y=3x+b$$
Since the point $(0,2)$ also lies in the tangent line $(l)$, we can apply that point to the equation of $(l)$ to find $b$. In detail, $$3\times0+b=2$$$$b=2$$
Therefore, the equation of the tangent line to the curve at $(0,2)$ is $$(l):y=3x+2$$
