Question

1.determine symetry
x^5-31x^3+150x=0

Answer 1

testing for symmetry for
y-axis change "x" to "-x"

x-axis change "y" to "-y"

origin line, change "x" and "y" to "-x" and "-y"

if after the change, you end up with the same original equation
then it has symmetry to that axis or line

for example... let's do the y-axis test
"x' to "-x"

so \(\large {
\underline{x^5-31x^3+150x=f(x)}\qquad \\ \quad \\({\color{brown}{ -x}})^5-31({\color{brown}{ -x}})^3+150({\color{brown}{ -x}})=f({\color{brown}{ -x}})
\\ \quad \\

\begin{cases}
(-x)^5\to -x\cdot -x\cdot-x\cdot-x\cdot-x\to &-x^5\\
(-x)^3\to -x\cdot-x\cdot-x\to &-x^3\\
(-x)\to &-x
\end{cases}\qquad thus
\\ \quad \\
-x^5-31(-x^3)+150(-x)=f(-x)\\ \quad \\\implies \underline{-x^5+31x^3-150x=f(-x)}
}\)

so.. notice the original equation
and f(-x), are not the same... thus, there's no symmetry over the y-axis

check for the x-axis and the origin

Answer 2

so the answer is no symetry?

Answer 3

@jdoe0001