(A) The Slope of the Secant Line

The slope of the secant line through the points $(2, f(2))$ and $(2 + h, f(2 + h))$ is given by the formula:

$m_{\text{secant}} = \frac{f(2 + h) - f(2)}{(2 + h) - 2}$

1. Find $f(2)$:

$f(2) = 2(2)^2 = 2 \times 4 = 8$

2. Find $f(2 + h)$:

$f(2 + h) = 2(2 + h)^2 = 2(4 + 4h + h^2) = 8 + 8h + 2h^2$

3. Calculate $m_{\text{secant}}$:

$m_{\text{secant}} = \frac{f(2 + h) - f(2)}{h} = \frac{(8 + 8h + 2h^2) - 8}{h} = \frac{8h + 2h^2}{h}$

Simplify:

$m_{\text{secant}} = \frac{h(8 + 2h)}{h} = 8 + 2h$

So, the slope of the secant line is:

$m_{\text{secant}} = 8 + 2h$

(B) The Slope of the Graph at $(2, f(2))$

The slope of the graph at a point is given by the derivative of $f(x)$ evaluated at that point.

1. Find $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^2) = 4x$

2. Evaluate $f'(2)$:

$f'(2) = 4(2) = 8$

So, the slope of the graph at $(2, f(2))$ is $8$.

(C) The Equation of the Tangent Line at $(2, f(2))$

The equation of the tangent line is given by:

$y - f(2) = f'(2)(x - 2)$

1. We know:

   • $f(2) = 8$
   • $f'(2) = 8$

2. Substitute these values:

$y - 8 = 8(x - 2)$

3. Simplify:

$y - 8 = 8x - 16 \quad \Rightarrow \quad y = 8x - 8$

So, the equation of the tangent line is:

$y = 8x - 8$

Would you like any further details or have any additional questions?

Follow-up Questions:

1. How do we interpret the meaning of the secant line in relation to the curve $f(x)$?
2. What is the geometric significance of the slope of the tangent line?
3. How can we find the equation of a normal line to $f(x)$ at $x = 2$?
4. What happens to the secant line slope as $h$ approaches zero?
5. Can we find the second derivative $f''(x)$ and interpret its meaning at $x = 2$?

Tip:

For finding tangent lines, always remember that the slope of the tangent line is the value of the derivative of the function at the given point.