Find the indicated quantities for \( f(x)=3 \mathrm{x}^{2} \). (A) The slope of the secant line through the points \( (2, f(2)) \) and \( (6, f(6)) \). (B) The slope of the secant line through the points \( (2, f(2)) \) and \( (2+\mathrm{h}, \mathrm{f}(2+\mathrm{h})), \mathrm{h} \neq 0 \) - (C) The slope of the graph at \( (2, f(2)) \) (A) The slope of the secant line through the points \( (2, f(2)) \) and \( (6, f(6)) \) is (Type an integer or a simplfied fraction.) (B) The slope of the secant line through the points \( (2, f(2)) \) and \( (2+h, f(2+h)), \mathrm{h} \neq 0 \), is (C) The slope of the graph at \( (2, f(2)) \) is \( \square \). (Type an integer or a simplitied fraction.)

Let's analyze each part step by step for the function \( f(x) = 3x^2 \). ### (A) Slope of the Secant Line through \( (2, f(2)) \) and \( (6, f(6)) \): The slope of a secant line between two points \( (x_1, f(x_1)) \) and \( (x_2, f(x_2)) \) is given by: \[ \text{Slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \] For \( x_1 = 2 \) and \( x_2 = 6 \): \[ f(2) = 3(2)^2 = 12 \] \[ f(6) = 3(6)^2 = 108 \] \[ \text{Slope} = \frac{108 - 12}{6 - 2} = \frac{96}{4} = 24 \] **Answer for (A):** \( 24 \) --- ### (B) Slope of the Secant Line through \( (2, f(2)) \) and \( (2+h, f(2+h)) \), \( h \neq 0 \): Using the same slope formula: \[ \text{Slope} = \frac{f(2+h) - f(2)}{(2+h) - 2} = \frac{f(2+h) - 12}{h} \] Calculate \( f(2+h) \): \[ f(2+h) = 3(2+h)^2 = 3(4 + 4h + h^2) = 12 + 12h + 3h^2 \] Now, plug this into the slope formula: \[ \text{Slope} = \frac{12 + 12h + 3h^2 - 12}{h} = \frac{12h + 3h^2}{h} = 12 + 3h \] **Answer for (B):** \( 12 + 3h \) --- ### (C) Slope of the Graph at \( (2, f(2)) \): The slope of the graph at a particular point is the derivative of the function at that point. First, find the derivative \( f'(x) \): \[ f(x) = 3x^2 \implies f'(x) = 6x \] Now, evaluate the derivative at \( x = 2 \): \[ f'(2) = 6 \times 2 = 12 \] **Answer for (C):** \( 12 \)