find an equation of the tangent line to the curve at the given point y=2e^x+x (0,2)

Question

Please help

Answer 1

Since the function  is  f(x) = 2 e^x +x,      the derivative , f' , is   2 e^x +1.   The derivative is the slope,

so that at the point (0,2), the slope (m) is   2 e⁰ + 1   =  3.

 

   Once the slope is known, the algebra 1  formula   y - y₁ = m (x - x₁)  is useful

 

    y - 2 =  3 ( x -0)      with rearrangement   y =   3x + 2