To sketch the graph of the linear function $f(x) = 9x + 8$, follow these steps:

Step 1: Identify the slope and y-intercept

The equation $f(x) = 9x + 8$ is in the slope-intercept form $y = mx + b$, where:

• $m = 9$ is the slope of the line.
• $b = 8$ is the y-intercept (the point where the line crosses the y-axis).

Step 2: Plot the y-intercept

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function: $f(0) = 9(0) + 8 = 8$ So, the point $(0, 8)$ is on the graph. Plot this point on the y-axis.

Step 3: Use the slope to find another point

The slope $m = 9$ means that for every 1 unit increase in $x$, $y$ increases by 9 units. Starting from the point $(0, 8)$, move 1 unit to the right (increasing $x$ by 1) and 9 units up (increasing $y$ by 9).

This gives another point: $x = 1, \quad f(1) = 9(1) + 8 = 17$ So, the point $(1, 17)$ is also on the graph. Plot this point.

Step 4: Draw the line

Now, draw a straight line passing through the points $(0, 8)$ and $(1, 17)$. This line is the graph of the function $f(x) = 9x + 8$.

Would you like me to generate a visual sketch of the graph for you?

---

Relative questions:

1. What is the slope-intercept form of a line and how does it help in graphing?
2. How would the graph change if the slope were negative, e.g., $f(x) = -9x + 8$?
3. What would be the graph's behavior if the y-intercept was 0?
4. How do you determine points to plot from a given linear equation?
5. How does the graph shift if the equation changes to $f(x) = 9x + 5$?

Tip: Always check the slope to understand how steep the line will be and whether it is increasing or decreasing.