Consider the function \( f(x)=3 x^{2}-5 x-4 \) and complete parts (a) through (c). (a) Find \( f(a+h) \). (b) Find \( \frac{f(a+h)-f(a)}{h}, h \neq 0 \). (c) Find the instantaneous rate of change of \( f \) when \( a=6 \).

Alright, I need to solve the problem involving the function \( f(x) = 3x^2 - 5x - 4 \) for parts (a), (b), and (c). Let's tackle each part step by step. **Part (a): Find \( f(a+h) \).** First, I need to find the value of the function \( f \) at \( x = a + h \). The function is given by: \[ f(x) = 3x^2 - 5x - 4 \] So, substituting \( x = a + h \) into the function: \[ f(a+h) = 3(a+h)^2 - 5(a+h) - 4 \] Now, I'll expand \( (a+h)^2 \): \[ (a+h)^2 = a^2 + 2ah + h^2 \] Substituting this back into the equation: \[ f(a+h) = 3(a^2 + 2ah + h^2) - 5(a+h) - 4 \] Distribute the 3: \[ f(a+h) = 3a^2 + 6ah + 3h^2 - 5a - 5h - 4 \] So, the expression for \( f(a+h) \) is: \[ f(a+h) = 3a^2 + 6ah + 3h^2 - 5a - 5h - 4 \] **Part (b): Find \( \frac{f(a+h)-f(a)}{h}, h \neq 0 \).** Next, I need to find the difference quotient, which is the average rate of change of the function over the interval from \( a \) to \( a+h \). The formula is: \[ \frac{f(a+h) - f(a)}{h} \] From part (a), I have \( f(a+h) \). Now, I need to find \( f(a) \): \[ f(a) = 3a^2 - 5a - 4 \] Subtracting \( f(a) \) from \( f(a+h) \): \[ f(a+h) - f(a) = (3a^2 + 6ah + 3h^2 - 5a - 5h - 4) - (3a^2 - 5a - 4) \] Simplifying: \[ f(a+h) - f(a) = 3a^2 + 6ah + 3h^2 - 5a - 5h - 4 - 3a^2 + 5a + 4 \] Combine like terms: \[ f(a+h) - f(a) = 6ah + 3h^2 - 5h \] Now, divide by \( h \): \[ \frac{f(a+h) - f(a)}{h} = \frac{6ah + 3h^2 - 5h}{h} \] Factor out \( h \) from the numerator: \[ \frac{f(a+h) - f(a)}{h} = \frac{h(6a + 3h - 5)}{h} \] Cancel out \( h \) (since \( h \neq 0 \)): \[ \frac{f(a+h) - f(a)}{h} = 6a + 3h - 5 \] So, the difference quotient is: \[ \frac{f(a+h) - f(a)}{h} = 6a + 3h - 5 \] **Part (c): Find the instantaneous rate of change of \( f \) when \( a=6 \).** The instantaneous rate of change of a function at a point is given by the derivative of the function at that point. From part (b), the difference quotient is: \[ \frac{f(a+h) - f(a)}{h} = 6a + 3h - 5 \] To find the derivative, take the limit as \( h \) approaches 0: \[ f'(a) = \lim_{h \to 0} (6a + 3h - 5) \] Substituting \( a = 6 \): \[ f'(6) = 6(6) + 3(0) - 5 = 36 + 0 - 5 = 31 \] So, the instantaneous rate of change of \( f \) when \( a = 6 \) is: \[ f'(6) = 31 \]