Question

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=\frac{e^{x}}{4}-x ; a=0$$

Short answer

Answer: The equation of the tangent line is $$y = -\frac{3}{4}x + \frac{1}{4}$$.

Step-by-step solution

Find the derivative

To find the derivative of $$y = \frac{e^x}{4} - x$$, we'll derive each term separately. The derivative of $$\frac{e^x}{4}$$ is $$\frac{1}{4}e^x$$, and the derivative of $$-x$$ is $$-1$$. Therefore, putting these together, the derivative of the function is: $$y' = \frac{1}{4}e^x - 1$$

Evaluate the derivative at a = 0

Now that we have the derivative, let's evaluate it at the given point $$a = 0$$: $$y'(0) = \frac{1}{4}e^0 - 1 = \frac{1}{4}(1) - 1 = -\frac{3}{4}$$ The slope of the tangent line at the point $$a = 0$$ is $$-\frac{3}{4}$$.

Find the coordinates of the point

To find the coordinates of the point where $$a=0$$, we will plug $$x=0$$ into the original function: $$y = \frac{e^0}{4} - 0 = \frac{1}{4}$$ So the coordinates of the point are $$(0, \frac{1}{4})$$.

Use the point-slope formula to find the equation of the tangent line

Now that we have the slope ($$-\frac{3}{4}$$) and coordinates ($$(0, \frac{1}{4})$$) of the point at $$a = 0$$, we can use the point-slope formula to find the equation of the tangent line: $$y - y_{1} = m(x - x_{1})$$ Plugging in the values: $$y - \frac{1}{4} = -\frac{3}{4}(x - 0)$$ Simplifying, we get: $$y = -\frac{3}{4}x + \frac{1}{4}$$ This is the equation of the tangent line.

Graph the curve and tangent line

Finally, use a graphing utility to graph the original function $$y = \frac{e^x}{4} - x$$ and the tangent line $$y = -\frac{3}{4}x + \frac{1}{4}$$ on the same set of axes. The student should overlay the graphs to see the tangent line touching the curve at the point $$(0, \frac{1}{4})$$.