Let's start by finding the average rate of change of the function $h(x) = x^2 - 5x$ from $x = 3$ to $x = 5$.

Part (a): Average Rate of Change

The average rate of change of a function between two points $x = a$ and $x = b$ is given by:

$\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}$

For this problem:

• $a = 3$
• $b = 5$
• $h(x) = x^2 - 5x$

Let's calculate $h(3)$ and $h(5)$:

1. Calculate $h(3)$:

$h(3) = 3^2 - 5 \cdot 3 = 9 - 15 = -6$

2. Calculate $h(5)$:

$h(5) = 5^2 - 5 \cdot 5 = 25 - 25 = 0$

Now, plug these values into the average rate of change formula:

$\text{Average Rate of Change} = \frac{h(5) - h(3)}{5 - 3} = \frac{0 - (-6)}{5 - 3} = \frac{6}{2} = 3$

So, the average rate of change from $x = 3$ to $x = 5$ is 3.

Part (b): Equation of the Secant Line

The secant line passes through the points $(3, h(3))$ and $(5, h(5))$, which are $(3, -6)$ and $(5, 0)$. The slope of the secant line is the same as the average rate of change, which is 3.

To find the equation of the secant line, use the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Using point $(3, -6)$ and slope $m = 3$:

$y - (-6) = 3(x - 3)$

Simplify the equation:

$y + 6 = 3x - 9$

$y = 3x - 9 - 6$

$y = 3x - 15$

So, the equation of the secant line is $y = 3x - 15$.

Would you like more details or have any questions about these steps? Here are some related questions:

1. How do you find the instantaneous rate of change of a function at a point?
2. What is the difference between the average rate of change and the instantaneous rate of change?
3. How can you determine the concavity of a function based on its second derivative?
4. What are the properties of a secant line versus a tangent line?
5. How does the average rate of change relate to the derivative of a function?

Tip: The average rate of change of a function over an interval is the same as the slope of the secant line passing through the endpoints of the interval.