Question

which of the following polynomials has a graph with odd symmetry? a. 7x^4+13x^3-11 b. 5x^7+4x^6+3x^5+x c.13x^5 -1/2x^3+7x+3 d. 8x^5+7x^3-5x

Answer 1

D)
`f(x)=8x^5+7x^3-5x`

Explanation:

A function is odd if its graph is symmetric to the origin, which we can check this if
`f(-x)=-f(x)` is true:

Option A

`f(x)=7x^4+13x^3-11\\\\f(-x)=7(-x)^4+13(-x)^3-11\\\\f(-x)=7x^4-13x^3-11`

Since
`f(-x)\\eq-f(x)`, then the function does not have odd symmetry

Option B

`f(x)=5x^7+4x^6+3x^5+x\\\\f(-x)=5(-x)^7+4(-x)^6+3(-x)^5+x\\\\f(-x)=-5x^7+4x^6-3x^5+x`

Since
`f(-x)\\eq-f(x)`, then the function does not have odd symmetry

Option C

`f(x)=13x^5-(1)/(2)x^3+7x+3\\\\f(-x)=13(-x)^5-(1)/(2)(-x)^4+7(-x)+3\\ \\f(-x)=-13x^5-(1)/(2)x^4-7x+3`

Since
`f(-x)\\eq-f(x)`, then the function does not have odd symmetry

Option D

`f(x)=8x^5+7x^3-5x\\\\f(-x)=8(-x)^5+7(-x)^3-5(-x)\\\\f(-x)=-8x^5-7x^3+5x`

Since
`f(-x)=-f(x)`, then the function DOES have odd symmetry. You can also see that the function is odd because every term has an odd exponent.