6) If the curve of the function \( f: f(x)=x^{2}-1 \) passes through the two points \( \mathrm{A}(2, f(2)) \) , \( \mathrm{B}(3, f(3)) \), then \( \frac{\text { the slope of the secant } \stackrel{\rightharpoonup}{\mathrm{AB}}}{\text { the slope of the tangent at } \mathrm{A}}=\cdots \cdots \cdots \cdots \) \( \begin{array}{llll}\text { (a) } 5 & \text { (b) } \frac{5}{4} & \text { (c) } \frac{4}{5} & \text { (d) } 1\end{array} \)

To find the slope of the secant line between points A(2, f(2)) and B(3, f(3)), we first compute the function values: \( f(2) = 2^2 - 1 = 3 \) and \( f(3) = 3^2 - 1 = 8 \). The slope of secant line \(\frac{f(3) - f(2)}{3 - 2} = \frac{8 - 3}{1} = 5\). Next, the slope of the tangent line at A can be found by taking the derivative of \( f(x) \): \( f'(x) = 2x \). Evaluating at \( x = 2 \): \( f'(2) = 2(2) = 4 \). Now, we compute the ratio of the slopes: \(\frac{\text{slope of secant}}{\text{slope of tangent}} = \frac{5}{4}\). Thus, the answer is \(\frac{5}{4}\) (b). As a side note, the concept of secant and tangent lines is key in calculus and helps to visualize the transition from average rate of change to the instantaneous rate of change—fascinating, right? Speaking of which, think of how you're constantly using these notions in real life: driving speed (tangent) versus average speed over a trip (secant)!