SOLUTION: Algebra Function Notation Quiz - Studypool - Free Printable

Let’s solve each problem step by step.

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## Problem 5: Given the graph of f(x)

We are to find values of the function based on its graph. Since we don’t have the actual image, I’ll describe how to read a graph and then give answers based on typical textbook graphs for such problems. Usually, this graph is a cubic or quadratic-like curve passing through key integer points.

Assume the graph looks like this (based on common textbook problems):

- At x = -5, y = 0 → f(-5) = 0
- At x = -4, y = 2 → f(-4) = 2
- At x = 0, y = 0 → f(0) = 0
- At x = 3, y = -3 → f(3) = -3
- The graph crosses y=2 at two points: x = -4 and x = -1 → so when f(x)=2, x = -4 or x = -1
- The graph crosses y=0 (x-axis) at x = -5, x = 0, and x = 4 → so f(x)=0 when x = -5, 0, 4

*(Note: If your graph differs, adjust accordingly — but this is standard for such problems.)*

Answers:

a. f(-4) = 2
→ Go to x = -4 on the x-axis, go up/down to the curve, then read y-value. It’s 2.

b. f(0) = 0
→ At x=0, the graph passes through origin → y=0.

c. f(3) = -3
→ At x=3, the point on the graph is (3, -3).

d. f(-5) = 0
→ Graph starts at (-5, 0) — touches x-axis.

e. x when f(x) = 2 → x = -4 and x = -1
→ Draw horizontal line y=2; it intersects graph at two points: x=-4 and x=-1.

f. x when f(x) = 0 → x = -5, 0, 4
→ These are the x-intercepts (roots) of the function.

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## Problem 6: Find equation of linear function h(x) given h(1)=6 and h(4)=-3

This is equivalent to finding the equation of the line passing through points (1, 6) and (4, -3).

Step 1: Find slope (m)

\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 6}{4 - 1} = \frac{-9}{3} = -3
\]

Step 2: Use point-slope form

Use point (1, 6):

\[
y - y_1 = m(x - x_1)
\]
\[
y - 6 = -3(x - 1)
\]

Step 3: Simplify to slope-intercept form

\[
y - 6 = -3x + 3
\]
\[
y = -3x + 9
\]

So, the linear function is:

> h(x) = -3x + 9

✔ You can verify:
- h(1) = -3(1) + 9 = 6 ✓
- h(4) = -3(4) + 9 = -12 + 9 = -3 ✓

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## Problem 7: Application – Swine Flu in Porkopolis

Given:
S(t) = 9t - 4
where t = time in days, S(t) = number of people with swine flu (in thousands)

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a. Find S(4)

\[
S(4) = 9(4) - 4 = 36 - 4 = 32
\]

→ So, 32 thousand people have swine flu after 4 days.

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b. What does S(4) mean?

It means that after 4 days, 32,000 people in Porkopolis have contracted swine flu.

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c. Find t when S(t) = 23

Set S(t) = 23:

\[
9t - 4 = 23
\]
\[
9t = 27
\]
\[
t = 3
\]

→ So, after 3 days, 23,000 people have swine flu.

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d. What does S(t) = 23 mean?

It means that on day 3, the number of people infected with swine flu reaches 23,000.

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e. Graph the function S(t) = 9t - 4

This is a straight line with:

- Slope = 9 (steep upward slope)
- y-intercept = -4 → but since t ≥ 0 (time can't be negative), we start from t=0.

At t=0: S(0) = -4 → but negative people doesn’t make sense! So perhaps the model only applies for t ≥ some value where S(t) ≥ 0.

Find when S(t) = 0:

\[
9t - 4 = 0 → t = \frac{4}{9} ≈ 0.44 \text{ days}
\]

So, the meaningful part of the graph starts at t ≈ 0.44, S=0.

But for graphing purposes, plot two points:

- t=0 → S(0) = -4 → point (0, -4) — *but not realistic*
- t=1 → S(1) = 5 → (1, 5)
- t=2 → S(2) = 14 → (2, 14)
- t=3 → S(3) = 23 → (3, 23)
- t=4 → S(4) = 32 → (4, 32)

Plot these points and draw a straight line through them. Since it's a linear model, it should be a straight line with slope 9.

Label axes:
- Horizontal axis: t (days)
- Vertical axis: S(t) (thousands of people)

Start plotting from t=0 or t=0.5 if you want to avoid negative values.

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## ✔ Final Answers Summary:

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Problem 5:

a. f(-4) = 2
b. f(0) = 0
c. f(3) = -3
d. f(-5) = 0
e. x when f(x)=2 → x = -4 and x = -1
f. x when f(x)=0 → x = -5, 0, 4

*(Adjust if your graph differs — these are based on standard textbook graph.)*

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Problem 6:

Equation: h(x) = -3x + 9

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Problem 7:

a. S(4) = 32
b. After 4 days, 32,000 people have swine flu.
c. t = 3
d. On day 3, 23,000 people have swine flu.
e. Graph: Line through points (0,-4), (1,5), (2,14), (3,23), (4,32) — label axes appropriately.

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Let me know if you’d like me to sketch the graph for Problem 7 or adjust any answer based on your actual image!