91Ó°ÊÓ

First, we find the derivative of the given function \(y(x)=\frac{e^x}{4}-x\) with respect to \(x\). Use the properties of derivatives: \(y'(x) = \frac{d}{dx}[\frac{e^x}{4}-x]\) Since the derivative of a sum/difference is the sum/difference of the derivatives, we can rewrite the expression: \(y'(x) = \frac{d}{dx}[\frac{e^x}{4}]-\frac{d}{dx}[x]\) Now, find the derivatives of each term: \(y'(x)=\frac{1}{4}e^x - 1\)

Now, we need to find the derivative at \(x=a=0\). Substitute \(x=0\) into the expression we found: \(y'(0)=\frac{1}{4}e^0-1\) We know that \(e^0 = 1\), so we have: \(y'(0)=\frac{1}{4}-1\) After simplifying, \(y'(0)=-\frac{3}{4}\)

To find the point on the curve where \(x=a=0\), substitute the value of \(a\) into the original function \(y(x)=\frac{e^x}{4}-x\): \(y(0)=\frac{e^0}{4}-0\) Since \(e^0=1\), we have: \(y(0)=\frac{1}{4}\) So, the point on the curve is \((0,\frac{1}{4})\).

We know the slope of the tangent line from step 2, which is \(-\frac{3}{4}\), and we have the point \((0, \frac{1}{4})\) on the curve from step 3. Now we can use the point-slope form of a line to find the equation of the tangent line: \(y-y_1=m(x-x_1)\) Substitute the values: \(y-\frac{1}{4}=-\frac{3}{4}(x-0)\) Then, simplify the equation: \(y=\frac{1}{4}-\frac{3}{4}x\) The equation of the tangent line at \(x=a=0\) is \(y=\frac{1}{4}-\frac{3}{4}x\). To graph the curve and the tangent line, use a graphing utility like Desmos or a graphing calculator. Enter the function \(y(x)=\frac{e^x}{4}-x\) and the tangent line equation \(y=\frac{1}{4}-\frac{3}{4}x\). You will see the curve and the tangent line on the same set of axes.