Worksheet 1.5 - Analyzing Graphs of Functions, featuring a semicircular graph to determine domain and range.

Let's go through each problem step by step and solve them with explanations.

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Problem 1: Find the domain and range of the function.

Graph Description:
The graph is a semicircle (top half) centered at the origin, opening downward, with endpoints at $ x = -4 $ and $ x = 4 $. The highest point is at $ y = 4 $, and it touches the x-axis at $ x = -4 $ and $ x = 4 $.

- The domain is the set of all possible $ x $-values.
- The range is the set of all possible $ y $-values.

From the graph:
- The curve starts at $ x = -4 $ and ends at $ x = 4 $ → Domain: $ [-4, 4] $
- The $ y $-values go from $ y = 0 $ (at the endpoints) up to $ y = 4 $ (at the top) → Range: $ [0, 4] $

✔ Answer:
- Domain: $ [-4, 4] $
- Range: $ [0, 4] $

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Problem 2: Use the graph to find the indicated function values.

We are given a graph of $ y = f(x) $ that looks like a wavy cubic-like function passing through several points.

Let’s examine each value:

#### a) $ f(-2) $
- Look at $ x = -2 $ on the graph.
- Trace vertically up to the curve.
- The corresponding $ y $-value appears to be $ y = 2 $.

✔ $ f(-2) = 2 $

#### b) $ f(-1) $
- At $ x = -1 $, the graph is at its peak just before dropping down.
- The $ y $-value is approximately $ y = 3 $.

✔ $ f(-1) = 3 $

#### c) $ f\left(\frac{1}{2}\right) $
- $ x = 0.5 $ is halfway between 0 and 1.
- At this point, the graph is decreasing from the peak at $ x = 0 $.
- The $ y $-value appears to be about $ y = 1 $.

✔ $ f\left(\frac{1}{2}\right) = 1 $

#### d) $ f(1) $
- At $ x = 1 $, the graph crosses the x-axis → $ y = 0 $

✔ $ f(1) = 0 $

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Problem 3: Use the Vertical Line Test to determine whether $ y $ is a function of $ x $.

Given: $ y = \frac{1}{2}x^2 $

- This is a parabola opening upward, vertex at the origin.
- For every $ x $, there is exactly one $ y $.
- Any vertical line will intersect the graph in at most one point.

✔ Yes, $ y $ is a function of $ x $.

> Vertical Line Test: If any vertical line intersects the graph more than once, it's not a function. Here, no such case occurs.

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Problem 4: Use the Vertical Line Test to determine whether $ y $ is a function of $ x $.

Given: $ x - y^2 = 1 $ → rewrite as $ x = y^2 + 1 $

This is a parabola opening to the right.

- For example, when $ x = 2 $, $ y^2 = 1 $ → $ y = \pm 1 $
- So, one $ x $-value corresponds to two $ y $-values.

Try drawing vertical lines:
- A vertical line at $ x = 2 $ intersects the graph at two points: $ (2,1) $ and $ (2,-1) $

✘ No, $ y $ is not a function of $ x $.

> Because some vertical lines intersect the graph twice, it fails the Vertical Line Test.

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✔ Final Answers:

#### 1.
- Domain: $ [-4, 4] $
- Range: $ [0, 4] $

#### 2.
a) $ f(-2) = 2 $
b) $ f(-1) = 3 $
c) $ f\left(\frac{1}{2}\right) = 1 $
d) $ f(1) = 0 $

#### 3.
Yes, $ y = \frac{1}{2}x^2 $ is a function of $ x $.

#### 4.
No, $ x - y^2 = 1 $ is not a function of $ x $.

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