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Let's solve the problem step by step:

### (a) Find the average rate of change from \( -6 \) to \( 8 \) for \( g(x) = 7x^2 - 5 \).

The formula for the average rate of change of a function \( g(x) \) over an interval \( [a, b] \) is:
\[
\text{Average rate of change} = \frac{g(b) - g(a)}{b - a}
\]
Here, \( a = -6 \) and \( b = 8 \).

#### Step 1: Compute \( g(-6) \)
\[
g(-6) = 7(-6)^2 - 5 = 7(36) - 5 = 252 - 5 = 247
\]

#### Step 2: Compute \( g(8) \)
\[
g(8) = 7(8)^2 - 5 = 7(64) - 5 = 448 - 5 = 443
\]

#### Step 3: Calculate the average rate of change
\[
\text{Average rate of change} = \frac{g(8) - g(-6)}{8 - (-6)} = \frac{443 - 247}{8 + 6} = \frac{196}{14} = 14
\]
Thus, the average rate of change is **14**.

---

### (b) Find an equation of the secant line containing \( (-6, g(-6)) \) and \( (8, g(8)) \).

The equation of a line in slope-intercept form is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

#### Step 1: Find the slope \( m \)
The slope of the secant line is the same as the average rate of change, which we found to be \( 14 \).

#### Step 2: Use the point-slope form to find the equation
We use the point-slope form of a line equation:
\[
y - y_1 = m(x - x_1)
\]
We can use one of the points, say \( (-6, 247) \), and the slope \( m = 14 \):
\[
y - 247 = 14(x + 6)
\]

#### Step 3: Simplify to slope-intercept form
\[
y - 247 = 14x + 84
\]
\[
y = 14x + 84 + 247
\]
\[
y = 14x + 331
\]
Thus, the equation of the secant line is \( y = 14x + 331 \).

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Do you have any questions, or would you like further details?

Here are 5 related questions to explore:
1. How does the average rate of change relate to the derivative?
2. Can the slope of a secant line ever be negative for a quadratic function?
3. What happens if the points for the secant line are closer together?
4. How can you verify that the equation of the secant line is correct?
5. How do you find the equation of a tangent line at a given point?

**Tip:** The secant line represents the average slope between two points on a curve, while a tangent line represents the instantaneous slope at a single point.

Let g(x) = 7x^2 - 5. (a) Find the average rate of change from -6 to 8. (b) Find an equation of the secant line containing (-6, g(-6)) and (8, g(8)).

Learn how to find the average rate of change and the equation of a secant line for the quadratic function g(x) = 7x^2 - 5 over the interval from -6 to 8. This problem covers key algebraic concepts including the average rate of change and the slope of the secant line, suitable for students in Grades 9-12.

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