(a) For the linear function \( f(x)=x+5 \), evaluate \( f(-2) \) and \( f(4) \).

(b) Graph \( f \). How can the graph of \( f \) be used to determine the zero of \( f \) ?

(c) Find the zero of \( f \).

(a) \( f(-2)= \) \( \square \)

(Type an integer or a fraction.)

Let's work through the problem step by step.

### Part (a): Evaluate \( f(-2) \) and \( f(4) \).

The linear function is given by \( f(x) = x + 5 \).

1. **Evaluate \( f(-2) \):**
   \[
   f(-2) = (-2) + 5 = 3
   \]

2. **Evaluate \( f(4) \):**
   \[
   f(4) = 4 + 5 = 9
   \]

So:

- \( f(-2) = 3 \)
- \( f(4) = 9 \)

### Part (b): Graph \( f(x) = x + 5 \)

The graph of \( f(x) = x + 5 \) is a straight line with a slope of 1 and a y-intercept at \( (0, 5) \).

### Using the Graph to Determine the Zero of \( f \):

The zero of the function \( f(x) \) is the value of \( x \) where \( f(x) = 0 \). On the graph, this is the point where the line crosses the x-axis.

### Part (c): Find the Zero of \( f(x) = x + 5 \)

To find the zero, set \( f(x) = 0 \) and solve for \( x \):

\[
x + 5 = 0
\]
\[
x = -5
\]

So, the zero of the function is \( x = -5 \).

### Summary:

- (a) \( f(-2) = 3 \)
- (b) The graph of \( f(x) = x + 5 \) crosses the x-axis at \( x = -5 \).
- (c) The zero of \( f(x) \) is \( x = -5 \).

Would you like further details on any part of this solution?

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Here are five related questions you might explore:

1. What is the slope of the function \( f(x) = x + 5 \)?
2. How does changing the value of the constant term (e.g., \( x + 3 \) instead of \( x + 5 \)) affect the graph?
3. What is the y-intercept of the function \( f(x) = x + 5 \)?
4. How would you find the zero of the function \( f(x) = 2x + 7 \)?
5. What does the zero of a function represent in a real-world context?

**Tip:** The slope of a linear function tells you how steep the line is, while the y-intercept indicates where the line crosses the y-axis.

Explore how to evaluate f(x) = x + 5 for specific values and find the zero of the function using graphing methods. This page provides step-by-step solutions and explanations suitable for Grades 7-9 students.

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