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Math4Life
Group Title
The graph of the equation y = x2 – 3 is symmetric with respect to which of the following?
A. the x–axis
B. the y–axis
C. the origin
D. none of these
 one year ago
 one year ago
Math4Life Group Title
The graph of the equation y = x2 – 3 is symmetric with respect to which of the following? A. the x–axis B. the y–axis C. the origin D. none of these
 one year ago
 one year ago

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mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
option a
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
@mayankdevnani and how do you know that?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
look at the exponents on the right side...what exponents do you see?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
^by exponents i mean exponents of the variable/s
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
i assume @Math4Life is currently offline.....
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
Sorry I was afk for 1 moment, im back though, sorry
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
so what exponents do you see on the right side? (again i mean exponents of the variable/s)
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
just exponents...not the terms
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
x2 and 3?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
i think you are confused.... dw:1350310291294:dw
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
oops my bad, so ^2
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
close.... remember this: when you have a constant (which are whole numbers without variables) the exponent of their variable is 0 so the exponents of the variables here are 2 and 0 got it?
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
I got it so far
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
good. now here comes the basic part. are 2 and 0 even, odd or neither?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
good. now, when *all* the exponents of the variables *on the right side* are even, then it is automatically symmetric to the yaxis. if the exponents of the variables *on the right side* are *all* odd, then it is symmetric to the yaxis. if the exponents of the left side and right side have the same parity (parity means that the exponents of the left side and right side are either BOTH odd or BOTH even), then it is symmetric to the origin. If neither, then neither. did you follow that?
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
Ya, i understand now, thanks!
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
here are some examples to guide you: even function: \[y = x^4 + x^2 + 3\] \[y = x^2\] \[y = 3\] odd function \[y= x^3 + x\] \[y = x^5 + x\] \[y = x^5 + x^3 + x\] both \[y = x\] \[y^2 = x^2\] \[y^3 = x^5 + x^3\] neither \[y = x^2 + x + 1\] \[y + x^3 + 1\] \[y + x^3 + x^2\]
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
those last two parts should be \[y = x^3 + 1\] \[y = x^3 + x^2\]
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
in case you get confused..you can refer to these examples
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
by the way.. "both" means symmetric to origin
 one year ago

Math4Life Group TitleBest ResponseYou've already chosen the best response.1
Thanks so much!
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
welcome
 one year ago
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