Question

math nation fl algebra 2 -> polynomial functions -> even, odd, o stepping stones
complete the statem
the graph of $f(x)$ is
symmetric about the y - axis
symmetric about the x - axis
symmetric about the origin
not symmetric

Explanation:

Step1: Recall symmetry definitions

• A graph is symmetric about the \( y \)-axis if for every point \( (x,y) \) on the graph, \( (-x,y) \) is also on the graph.
• A graph is symmetric about the \( x \)-axis if for every point \( (x,y) \) on the graph, \( (x,-y) \) is also on the graph.
• A graph is symmetric about the origin if for every point \( (x,y) \) on the graph, \( (-x,-y) \) is also on the graph.

Step2: Analyze the given graph

Looking at the graph, we can see that if we take a point \( (x,y) \) on the right side of the \( y \)-axis (e.g., \( (1,1) \)), the point \( (-1,1) \) is also on the graph. This shows that the graph is symmetric with respect to the \( y \)-axis.

To check for \( x \)-axis symmetry, if we take a point \( (x,y) \), say \( (0,0) \), \( (0,-0)=(0,0) \) is on the graph, but for a point like \( (1,1) \), \( (1,-1) \) is not on the graph, so it's not symmetric about the \( x \)-axis.

For origin symmetry, if \( (x,y) \) is on the graph, \( (-x,-y) \) should be. Taking \( (1,1) \), \( (-1,-1) \) is not on the graph, so it's not symmetric about the origin.

Answer:

symmetric about the y - axis