Question

question 13 (1 point)

the point \\((-5, -5)\\) is on a graph of an odd function. what other point(s) must be on the graph of the function?

\\((5, 5)\\)

\\((5, -5)\\)

\\((0, 0)\\) & \\((-5, 5)\\)

\\((0, 0)\\) & \\((5, 5)\\)

\\((0, 0)\\) & \\((5, -5)\\)

\\((0, 0)\\)

Explanation:

🆕 New Concept Discovered: Odd Functions
Symmetry about the origin

Step 1: Definition of an Odd Function

An odd function \( f(x) \) satisfies the algebraic condition:

$$ f(-x) = -f(x) $$

In terms of coordinates, if a point \( (x, y) \) lies on the graph of an odd function, then the point \( (-x, -y) \) must also lie on the graph. This represents a reflection through the origin.

Step 2: Apply the Symmetry Rule

We are given that the point \( (-5, -5) \) is on the graph. Here, \( x = -5 \) and \( y = -5 \).

Applying the rule for odd functions:

• Negate the \( x \)-coordinate: \( -x = -(-5) = 5 \)
• Negate the \( y \)-coordinate: \( -y = -(-5) = 5 \)

Therefore, the point \( (5, 5) \) must also be on the graph.

Step 3: Check the Origin (0, 0)

For any odd function defined at \( x = 0 \), we must have:

$$ f(-0) = -f(0) \implies f(0) = -f(0) \implies 2f(0) = 0 \implies f(0) = 0 $$

This means that if \( x = 0 \) is in the domain of the odd function, the graph must pass through \( (0, 0) \). However, the problem does not state that the function is defined at \( x = 0 \). An odd function like \( f(x) = \frac{1}{x} \) is odd but does not contain the point \( (0, 0) \) because it is undefined at \( x = 0 \).

Therefore, we cannot guarantee that \( (0, 0) \) is on the graph. The only point that must be on the graph is \( (5, 5) \).

Answer:

\( (5, 5) \)