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Explain how this function is symmetric?

There is a multiple choice math problem in which I must choose the function/equation which is NOT symmetric along the origin, x-axis, or y-axis (any of them). The exact wording is: "Which of the following is NOT symmetric with respect to either the origin, the x-axis or the y-axis?"

I put each into my graphing calculator and found that 3/5 were indeed symmetric. However, I got stuck between:

y = x^2 + 2x + 1 (which is symmetric but across x= -1)

y = -x + 1 (which would be symmetric over the perpendicular slope, which is NOT the origin, x-axis, or y-axis)

The book claims that answer C, y = x^2 + 2x + 1, is the answer. So how is y = -x + 1 symmetric? I don't think it is.

The book says to substitute (-x, -y) to see if it is symmetric at the origin. If you replace x and y with this, you should come out with the same solution.

-y = -(-x) + 1

-y = x + 1

y = -x -1

So it's not symmetric at the origin.

Replace with (x, -y) to see if it's symmetric at the x-axis.

-y = -x + 1

y = x - 1

So it's not symmetric at the x-axis.

Replace with (-x, y) to see if it's symmetric at the y-axis

y = -(-x) + 1

y = x + 1

So it's not symmetric there either.

Am I missing something? I'm not a genius at math but I fail to see my error. If you graph it, it is definitely NOT symmetric at any of those spots, as it is a straight line which cuts through both the x-axis and y-axis at 1 (0,1) and (1,0).

Please confirm my belief that the book is missing something or correct me and tell me where I am erring. It may be eliminating this answer because it is symmetric with the line y = x, but that was not part of the question. Thank you!

2 Answers

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  • ted s
    Lv 7
    1 decade ago
    Favorite Answer

    GOOD for you , neither are symmetric about either axis or origin

  • ?
    Lv 4
    4 years ago

    There are a team of distinctive approaches of pointing out it, yet an equation is seen a function, this is the y term is written as f(x), IFF for each fee the enter has there is largely one output, this is comparable to saying that for any x there could be basically one y, So any 2nd order equations of the from y squared = x isn't a function. it relatively is clearer in case you rewrite the equation as sq. root of y = x If x is 0 then y is 0 this is one to a minimum of one, yet while x is a million y could be -a million or +a million if x is two y could be -4 or +4. for each fee of x there at the instant are 2 values of y, so via definition it is not a function. you may make it a function via saying that the effective sq. root of y = x as one equation and the adverse sq. root of y = x for yet another equation, which might conceal the unique equation area yet shop the only to a minimum of one correspondence.

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