Answer
The equation of the tangent line at the given point is $y=3x+1$
Work Step by Step
$y=1+2x+xe^{x}$ $;$ $a=0$
First, evaluate the derivative of the given expression. Use the product rule for the term $xe^{x}$:
$y'=0+2+x(e^{x})'+(x)'e^{x}=...$
$...=2+xe^{x}+e^{x}=2+e^{x}(x+1)$
Substitute $x$ by $a=0$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{\text{tan}}=2+e^{0}(0+1)=2+1=3$
Substitute $x$ by $a=0$ in the original expression to obtain the $y$-coordinate of the point given:
$y=1+2(0)+(0)e^{0}=1$
The point is $(0,1)$
The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-1=3(x-0)$
$y-1=3x$
$y=3x+1$
The graph of both the function and the tangent line are shown the answer section.
