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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:
\[
y = 6x - 3
\]

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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.

Find the indicated quantities for f(x) = 3x^2. (A) The slope of the secant line through the points (1, f(1)) and (1 + h, f(1 + h)), h ≠ 0. (B) The slope of the graph at (1, f(1)). (C) The equation of the tangent line at (1, f(1)).

Learn how to find the slope of the secant line, the slope of the graph at (1, f(1)), and the equation of the tangent line for the function f(x) = 3x^2. This involves applying derivative concepts and understanding secant and tangent lines, suitable for students in Grades 10-12.

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