Question

state the symmetry of the function

• the function has line symmetry about the y-axis
• the function has rotational symmetry about the origin
• the function does not have any symmetry
• the function has line symmetry about the x-axis

Explanation:

To determine the symmetry of the function (a parabola opening upwards with vertex at the origin), we analyze each option:

• Line symmetry about the y - axis: A function has y - axis symmetry if for every point \((x,y)\) on the graph, \((-x,y)\) is also on the graph. For this parabola (likely \(y = x^{2}\) or similar), if we take a point \((x,y)\), substituting \(-x\) gives \(y=(-x)^{2}=x^{2}\), so \((-x,y)\) is on the graph. So it is symmetric about the y - axis.
• Rotational symmetry about the origin: A function has rotational symmetry about the origin if for every point \((x,y)\) on the graph, \((-x,-y)\) is also on the graph. For \(y = x^{2}\), if we take \((x,y)=(1,1)\), \((-x,-y)=(-1, - 1)\), and \(-1

eq(-1)^{2}\), so \((-1,-1)\) is not on the graph. So no rotational symmetry about the origin.

• The function does not have any symmetry: We already saw it has y - axis symmetry, so this is false.
• Line symmetry about the x - axis: A function has x - axis symmetry if for every point \((x,y)\) on the graph, \((x,-y)\) is also on the graph. For \(y = x^{2}\), if we take \((x,y)=(1,1)\), \((x,-y)=(1,-1)\), and \(-1

eq1^{2}\), so \((1,-1)\) is not on the graph. So no x - axis symmetry.

Answer:

The function has line symmetry about the y - axis