Answer
The equation of the tangent line is $y=-\dfrac{3}{4}x+\dfrac{1}{4}$
Work Step by Step
$y=\dfrac{e^{x}}{4}-x$ $;$ $a=0$
First, evaluate the derivative of the given expression:
$y'=\dfrac{e^{x}}{4}-1$
Substitute $x$ by $a=0$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{tan}=\dfrac{e^{0}}{4}-1=\dfrac{1}{4}-1=-\dfrac{3}{4}$
Substitute $x$ by $a=0$ in the original expression to obtain the $y$-coordinate of the point given:
$y=\dfrac{e^{0}}{4}-0=\dfrac{1}{4}$
The point is $(0,\dfrac{1}{4})$
The slope of the tangent line and a point through which it passes are now 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:
$y-\dfrac{1}{4}=-\dfrac{3}{4}(x-0)$
$y-\dfrac{1}{4}=-\dfrac{3}{4}x$
$y=-\dfrac{3}{4}x+\dfrac{1}{4}$
The graph of both the function and the tangent line are shown in the answer section.