Let's break down the problem and solve it step by step.

We are given the function $f(x) = 3x^2$, and we need to find:

(A) The slope of the secant line through the points $(1, f(1))$ and $(1 + h, f(1 + h))$, where $h \neq 0$.

The slope of the secant line is given by: $m_{\text{secant}} = \frac{f(1 + h) - f(1)}{(1 + h) - 1}$ First, calculate $f(1)$ and $f(1 + h)$: $f(1) = 3(1^2) = 3$ $f(1 + h) = 3(1 + h)^2 = 3(1 + 2h + h^2) = 3 + 6h + 3h^2$ Now substitute these into the formula for the secant slope: $m_{\text{secant}} = \frac{(3 + 6h + 3h^2) - 3}{h} = \frac{6h + 3h^2}{h} = 6 + 3h$ Thus, the slope of the secant line is: $m_{\text{secant}} = 6 + 3h$

(B) The slope of the graph at $(1, f(1))$.

The slope of the graph at a point is the derivative of the function evaluated at that point. First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(3x^2) = 6x$ Now, evaluate the derivative at $x = 1$: $f'(1) = 6(1) = 6$ So, the slope of the graph at $(1, f(1))$ is: $m = 6$

(C) The equation of the tangent line at $(1, f(1))$.

The equation of the tangent line is given by the point-slope form: $y - f(1) = m(x - 1)$ We already know that $f(1) = 3$ and $m = 6$, so the equation becomes: $y - 3 = 6(x - 1)$ Simplifying: $y = 6x - 6 + 3$ $y = 6x - 3$ Thus, the equation of the tangent line is: $y = 6x - 3$

Let me know if you'd like further clarifications or details!

Here are 5 related questions to think about:

1. What would the slope of the secant line be if $h \to 0$?
2. How would the derivative change if the function were $f(x) = 3x^3$?
3. What is the general formula for the slope of the tangent line at any point $x = a$?
4. How would you calculate the second derivative of $f(x) = 3x^2$?
5. What does the tangent line represent geometrically?

Tip:

The slope of the secant line gives an average rate of change, while the tangent line gives the instantaneous rate of change at a specific point.