# Which of the following equations has a graph that is symmetric with respect to the origin?

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A) y=x+1/x

B)y= -x^5+3x

C)y=x^4-2x^2+6

D)y=(x-1)^3+1

E)y=(x^2+1)^2-1

• Which of the following equations has a graph that is symmetric with respect to the origin?

In other words…which of the following equations has the property: f(-x) = -f(x)…so check each one for this property…

Evaluate f(-x) (in other terms…substitute -x for all x’s in the function) and -f(x)…(times the function by -1)….so….trying each options…

A) Assuming this is y = (x + 1)/x….

f(-x) = (-(x) + 1)/(-x)

f(-x) = -(-x +1)/x

f(-x) = (x – 1)/x

-f(x) = -(x + 1)/x

-f(x) = (-x – 1)/x

Nope it isn’t this one…

…if you had meant: y = x + (1/x)..

f(x) = x + (1/x)

f(-x) = (-x) + 1/(-x)

f(-x) = -x – 1/x

-f(x) = -(x + (1/x))

-f(x) = -x – (1/x)

IF you had meant y = x + (1/x)…then this one would be the answer…

B) y = -x^5 + 3x

f(x) = -x^5 + 3x

f(-x) = -(-x)^5 + 3(-x)

f(-x) = -(-1)x^5 – 3x

f(-x) = x^5 – 3x

-f(x) = -(-x^5 + 3x)

-f(x) = x^5 – 3x

So this one is also an odd function…so this is also an answer…since this property f(-x) = -f(x) works..

C) f(x) = x^4 – 2x^2 + 6

f(-x) = (-x)^4 – 2(-x)^2 + 6

f(-x) = x^4 – 2x^2 + 6

-f(x) = -(x^4 – 2x^2 + 6)

NOPE…f(-x) does not equal -f(x)….

D) y = (x – 1)^3 + 1

f(-x) = (-x – 1)^3 + 1

f(-x) = -(x + 1)^3 + 1

-f(x) = – ((x – 1)^3 + 1))

-f(x) = – (x – 1)^3 – 1

NOPE…the f(-x) does not equal -f(x)…

E) f(x) = (x^2 + 1)^2 – 1

f(-x) = ((-x)^2 +1)^2 – 1

f(-x) = (x^2 + 1)^2 – 1

-f(x) = -(x^2+ 1)^2 + 1

NOPE!

So the answer is B) y = -x^5 + 3x…assuming that A) is actually (x+1)/x….

B) is the only one with the property -f(x) = f(-x)…

• A. www.wolframalpha.com/input/?i=plot+y%3Dx%2B1%2Fx

B. www.wolframalpha.com/input/?i=plot+y%3D+-x^5%2B3x

C. www.wolframalpha.com/input/?i=plot+y%3Dx^4-2x^2%2B6

D. www.wolframalpha.com/input/?i=plot+y%3D%28x-1%29^3%2B1+

E. www.wolframalpha.com/input/?i=plot+y%3D%28x^2%2B1%29^2-1

A and B appear to be symmetric with respect to the origin. That is assuming youo wrote the equation the way you meant it.

Even exponents give plots that are symmetrical about a vertical line. Not always with respect to the origin.

www.mathwords.com/s/symmetric_origin.htm

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