Page 3 of 11 MATH 131 Final Exam June 29,2023 3. (10 points) Find the equation of the tangent line to the graph of $f(x)=\frac{2 x+3}{x+6}$ at the point at which $x=0$

Step 1 ：Given the function \(f(x)=\frac{2x+3}{x+6}\), we are asked to find the equation of the tangent line at the point where \(x=0\).

Step 2 ：The slope of the tangent line at a given point is the derivative of the function at that point. So, we first need to find the derivative of \(f(x)\).

Step 3 ：Using the quotient rule, the derivative of \(f(x)\) is \(f'(x) = \frac{2(x+6) - (2x+3)(1)}{(x+6)^2} = \frac{2}{x+6} - \frac{2x+3}{(x+6)^2}\).

Step 4 ：Substitute \(x=0\) into the derivative to find the slope of the tangent line, \(m = f'(0) = \frac{2}{6} - \frac{3}{36} = \frac{1}{4}\).

Step 5 ：Substitute \(x=0\) into the original function to find the y-coordinate of the point of tangency, \(y = f(0) = \frac{2(0)+3}{0+6} = \frac{1}{2}\).

Step 6 ：Finally, use the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\), to find the equation of the tangent line. Here, \(x_1 = 0\), \(y_1 = \frac{1}{2}\), and \(m = \frac{1}{4}\).

Step 7 ：Substitute these values into the point-slope form to get the equation of the tangent line, \(y - \frac{1}{2} = \frac{1}{4}(x - 0)\).

Step 8 ：Simplify this equation to get the final answer, \(y = \frac{1}{4}x + \frac{1}{2}\).

Step 9 ：\(\boxed{y = \frac{1}{4}x + \frac{1}{2}}\) is the equation of the tangent line to the graph of the function \(f(x)=\frac{2x+3}{x+6}\) at the point where \(x=0\).