The function \( f(x)=x^{2}-2 x+9 \) is defined for all real values of \( x \). (a) Find the value of \( x \) for which \( f^{\prime}(x)=8 \). (b) Hence, find the equation of the tangent to the curve of \( y=f(x) \) that has a slope of 8 . Give your answer in the form \( y=m x+c \).

To tackle this problem, let's go step by step. (a) First, we need to find \( f'(x) \), the derivative of the function \( f(x) = x^2 - 2x + 9 \). Calculating the derivative: \[ f'(x) = 2x - 2 \] Now, we set the derivative equal to 8: \[ 2x - 2 = 8 \] Solving for \( x \): \[ 2x = 10 \implies x = 5 \] So, the value of \( x \) for which \( f'(x) = 8 \) is \( x = 5 \). (b) Next, we need to find the equation of the tangent line to the curve at the point where \( x = 5 \) and with a slope of 8. First, we calculate \( f(5) \): \[ f(5) = 5^2 - 2 \cdot 5 + 9 = 25 - 10 + 9 = 24 \] Now, we have the point \( (5, 24) \) on the curve, and we know the slope \( m = 8 \). Using the point-slope form of a line, we can write the equation of the tangent line: \[ y - f(5) = m(x - 5) \] Substituting in the values: \[ y - 24 = 8(x - 5) \] Expanding and simplifying: \[ y - 24 = 8x - 40 \] \[ y = 8x - 40 + 24 \] \[ y = 8x - 16 \] Thus, the equation of the tangent line in the form \( y = mx + c \) is \[ \boxed{y = 8x - 16} \]