\(y=0\) everywhere on the \(x\) axis, so plug \(y=0\) into the equation and solve for \(x\) to find the \(x\) intercepts: \(x=-{2}\) and \(x={2}\text{.}\)
\(x=0\) everywhere on the \(y\) axis, so plug \(x=0\) into the equation and solve for \(y\) to find the \(y\) intercepts: \(y=-{2}\) and \(y={2}\text{.}\)
No, the graph is not symmetric about the \(x\)-axis. The point \((1,2)\) is on the graph but its reflection in the \(x\)-axis, \((1,-2)\text{,}\) is not.
Yes, the graph is symmetric about the \(y\)-axis. Replacing \(x\) with \(-x\) in the equation produces an equation \(\displaystyle{y=(-x)^4+(-x)^2=x^4+x^2}\) which is the same as the original equation. Thus \((x,y)\) satisfies the (original) equation if and only if its reflection in the \(y\)-axis \((-x,y)\) satisfies it.