Let ​h(x)equalsx squared minus 5 x.
​(a)  Find the average rate of change from 3 to 5.
​(b)  Find an equation of the secant line containing left parenthesis 3 comma h left parenthesis 3 right parenthesis right parenthesis and left parenthesis 5 comma h left parenthesis 5 right parenthesis right parenthesis .
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Part 1
​(a)  The average rate of change from 3 to 5 is

enter your response here.  ​(Simplify your​ answer.)

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.

Let h(x) equal x squared minus 5x. (a) Find the average rate of change from 3 to 5. (b) Find an equation of the secant line containing (3, h(3)) and (5, h(5)).

Learn how to find the average rate of change for the quadratic function h(x) = x^2 - 5x from x = 3 to x = 5. This problem also covers the steps to determine the equation of the secant line passing through points (3, h(3)) and (5, h(5)). Suitable for high school students in Grades 9-12.

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