Find the two \( x \)-intercepts of the function \( f \) and show that \( f^{\prime}(x)=0 \) at some point between the two \( x \)-intercepts. \[ f(x)=x^{2}-5 x-6 \] smaller \( x \)-value \( \quad(x, y)=(\square) \) larger \( x \)-value \( \quad(x, y)=(\square) \) Find a value of \( x \) such that \( f^{\prime}(x)=0 \). \( x=\square \)

To find the \( x \)-intercepts of the function \( f(x) = x^2 - 5x - 6 \), we set \( f(x) = 0 \): \[ x^2 - 5x - 6 = 0 \] This can be factored as: \[ (x - 6)(x + 1) = 0 \] Thus, the \( x \)-intercepts are \( x = 6 \) and \( x = -1 \). So, the smaller \( x \)-value is \( (x, y) = (-1, 0) \) and the larger \( x \)-value is \( (x, y) = (6, 0) \). Next, we find the derivative \( f'(x) \): \[ f'(x) = 2x - 5 \] Setting it equal to zero to find the critical point: \[ 2x - 5 = 0 \quad \Rightarrow \quad 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2} = 2.5 \] Thus, \( x = 2.5 \) is the value where \( f'(x) = 0 \). To summarize: Smaller \( x \)-value \( \quad(x, y) = (-1, 0) \) Larger \( x \)-value \( \quad(x, y) = (6, 0) \) Find a value of \( x \) such that \( f^{\prime}(x) = 0 \): \( x = 2.5 \) --- Did you know that quadratic functions, like the one we just explored, play a crucial role in physics, engineering, and economics? They're often used to model parabolic trajectories, such as the path of a baseball when hit or the optimization of profit and cost in business! In calculus, the location where the derivative is zero indicates a local maximum, minimum, or a saddle point. This is fundamental in optimization problems—think about locating the most cost-effective solution or maximizing resources! Keep playing with functions like quadratics; they show up everywhere in the real world!