## Calculus: Early Transcendentals (2nd Edition)

$x^{3}-y^{5}=0$ Check for symmetry about the $y$-axis by substituting $x$ by $-x$ in the given equation and simplifying: $(-x)^{3}-y^{5}=0$ $-x^{3}-y^{5}=0$ Since substituting $x$ by $-x$ in the equation didn't yield an equivalent equation, it is not symmetric about the $y$-axis. Check for symmetry about the $x$-axis by substituting $y$ by $-y$ in the given equation and simplifying: $x^{3}-(-y)^{5}=0$ $x^{3}+y^{5}=0$ Since substituting $y$ by $-y$ in the equation didn't yield an equivalent equation, it is not symmetric about the $x$-axis. Check for symmetry about the origin by substituting $x$ by $-x$ and $y$ by $-y$ in the given equation and simplifying: $(-x)^{3}-(-y)^{5}=0$ $-x^{3}+y^{5}=0$ Take out common factor $-1$ from the left side: $(-1)(x^{3}-y^{5})=0$ Take $-1$ to divide the right side: $x^{3}-y^{5}=\dfrac{0}{-1}$ $x^{3}-y^{5}=0$ Since the substitutions yielded an equivalent equation, it is symmetric about the origin. The graph is shown in the answer section