## Calculus: Early Transcendentals 8th Edition

The equation of the tangent line to the curve at $(0,2)$ is $$(l):y=3x+2$$
$$y=2e^x+x$$ 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 $$(l):y=3x+2$$