A) y=x+1/x
B)y= x^5+3x
C)y=x^42x^2+6
D)y=(x1)^3+1
E)y=(x^2+1)^21
2 Answers

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^42x^2%2B6
D. www.wolframalpha.com/input/?i=plot+y%3D%28x1%29^3%2B1+
E. www.wolframalpha.com/input/?i=plot+y%3D%28x^2%2B1%29^21
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