Question

math nation fl algebra 2 -> polynomial functions -> even, odd stepping stones
complete the statement
the graph of f(x) is
dropdown options: symmetric about the y - axis, symmetric about the x - axis, symmetric about the origin, not symmetric
graph of a parabola opening upwards with vertex at the origin

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

The graph shown is a parabola opening upwards with vertex at the origin \((0,0)\). Let's check points:

• Take \( x = 1 \), the \( y \)-value is \( 1 \) (approximate from the graph). Then for \( x=-1 \), the \( y \)-value is also \( 1 \).
• Take \( x = 2 \), the \( y \)-value is \( 4 \) (approximate), and for \( x = -2 \), the \( y \)-value is also \( 4 \).

This shows that for every \( x \), \( f(-x)=f(x) \), which means the graph is symmetric about the \( y \)-axis. It is not symmetric about the \( x \)-axis (since if we take a point \((1,1)\), \((1,-1)\) is not on the graph) and not symmetric about the origin (since \((-1,1)\) and \((1,-1)\) would need to be on the graph, but \((1,-1)\) is not).

Answer:

symmetric about the y - axis