Let's go through each problem:

Problem 4

Function given:
$f(x) = \frac{5 - x}{3}$

This can be rewritten in slope-intercept form $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

1. Rewrite the function:
   Rewrite $f(x) = \frac{5 - x}{3}$ by dividing each term in the numerator by 3: $f(x) = -\frac{1}{3}x + \frac{5}{3}$

2. Identify the slope and y-intercept:

   • Slope $m$: $-\frac{1}{3}$
   • Y-intercept $b$: $\frac{5}{3}$

3. Sketch the graph (conceptual):
   The graph is a line with a slope of $-\frac{1}{3}$, meaning it decreases by $\frac{1}{3}$ unit vertically for every 1 unit increase horizontally. It crosses the y-axis at $y = \frac{5}{3}$.

Problem 5

Goal: Find $f(x)$ if it is a linear function with the following properties:

1. Slope = -5 and $f\left(\frac{1}{4}\right) = 9$
2. $f(-2) = 1$ and $f(-4) = -3$

Let's analyze each property to form the function $f(x)$.

Approach for Part (i):

Since the slope $m = -5$, we can write the general form of $f(x)$ as: $f(x) = -5x + b$ where $b$ is the y-intercept.

Using the condition $f\left(\frac{1}{4}\right) = 9$: $f\left(\frac{1}{4}\right) = -5\left(\frac{1}{4}\right) + b = 9$ Solving this: $-\frac{5}{4} + b = 9$ $b = 9 + \frac{5}{4} = \frac{36 + 5}{4} = \frac{41}{4}$ So, the function is: $f(x) = -5x + \frac{41}{4}$

Approach for Part (ii):

Verify with the points $f(-2) = 1$ and $f(-4) = -3$ to confirm that this function is correct.

For $f(-2)$: $f(-2) = -5(-2) + \frac{41}{4} = 10 + \frac{41}{4} = \frac{40 + 41}{4} = \frac{81}{4} \neq 1$ There seems to be an inconsistency with this part. It appears these values may suggest a different function or could involve a constraint mismatch.

Would you like further clarification on these solutions or to delve into why the values might not match in Problem 5 (ii)?

Follow-Up Questions:

1. How would you convert a linear function to slope-intercept form from different starting forms?
2. Why is it useful to know both the slope and y-intercept in graphing a linear function?
3. How does the slope affect the angle and direction of a line on the graph?
4. Can there be a linear function that satisfies two different pairs of points if the slope is set?
5. What does it mean if two points do not match a given slope in a linear context?

Tip:

In linear functions, always verify constraints by substituting back values; mismatched constraints may indicate no solution with the given slope.