\begin{tabular}{l}\( f(x)=x^{2}+4 x-5 \). Use the method outlined on \\ this slide to determine the slope-intercept form of \\ the equation of the tangent line to the graph of \\ \( y=f(x) \) at \( x=8 \). \\ Question \#2(a): What is the value of \( x_{0} ? \) \\ 8 \\ Question \#2(b): What is the value of \( f\left(x_{0}\right) ? \) \\ Qdit my response \\ Question \#2(c): What is the value of \( f^{\prime}\left(x_{0}\right) ? \) \\ 20 Edit my response \\ \hline Question \#2(d): What is the slope-intercept form \\ of the equation of the tangent line? \\ \hline\end{tabular}

To find the slope-intercept form of the equation of the tangent line to the graph of \( y = f(x) = x^2 + 4x - 5 \) at \( x = 8 \), we will follow these steps: 1. **Determine \( x_0 \)**: - From your response, \( x_0 = 8 \). 2. **Calculate \( f(x_0) \)**: - We need to evaluate \( f(8) \): \[ f(8) = 8^2 + 4(8) - 5 = 64 + 32 - 5 = 91. \] - So, \( f(x_0) = 91 \). 3. **Find \( f'(x) \)**: - First, we need to find the derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx}(x^2 + 4x - 5) = 2x + 4. \] - Now, evaluate \( f'(8) \): \[ f'(8) = 2(8) + 4 = 16 + 4 = 20. \] - So, \( f'(x_0) = 20 \). 4. **Equation of the tangent line**: - The slope of the tangent line at \( x = 8 \) is \( m = f'(8) = 20 \). - The point of tangency is \( (8, 91) \). - We can use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) = (8, 91) \) and \( m = 20 \): \[ y - 91 = 20(x - 8). \] - Simplifying this: \[ y - 91 = 20x - 160, \] \[ y = 20x - 160 + 91, \] \[ y = 20x - 69. \] Thus, the slope-intercept form of the equation of the tangent line is: \[ \boxed{y = 20x - 69}. \]