Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

xy=2 Is this? a)symmetry at x-axis,y-axis, and/or at the origin

Precalculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

rewrite as y = 2/x... recognize this?
yes
also if f(-x) = f(x) then it is symmetric about y. if f(-x) = -f(x) it is symmetric about the origin. if f(x) = y or -y then it is symmetric about x.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

not or, and.
that is, x: -> y and x -> -y then it is not a function but is symmetric about the x-axis.
trying to keep up but im sure y=2/x is a function and i wanted to know if any of the combinations of symmetry x-axis,y-axis,origin(or just at least one of these) are possible
learn to do this: \[f(x) = \frac{ 2 }{ x }\Rightarrow f(-x) = \frac{ 2 }{ -x }=-\frac{ 2 }{ x }\Rightarrow f(-x) = -f(x)\] so it's symmetric about the origin. since it's a function. it can't be symmetric about the x-axis.

Not the answer you are looking for?

Search for more explanations.

Ask your own question