Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Pre Calc. Can anyone help me?

Mathematics
See more answers at brainly.com
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

What's the question?
"Can anyone help me?"
YES! :D

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

1 Attachment
\[f(x) = \frac{ 2 }{ x^2-2x-3 }=\frac{ 2 }{ (x+1)(x-3) }\]
okay
That means the x cannot be -1, 3 so the domain (-infi, -1)U(-1, 3)U(3, infi)
or x <-1, -13
okay
for x-intercept: set f(x)=0, then: \[0=\frac{ 2 }{ (x+1)(x-3) }\] since no x exist that satisfy the above equation, there is no x-intercept
for y-intercept: set x = 0, then: \[f(0) = \frac{ 2 }{ (0+1)(0-3) }=\frac{ -2 }{ 3 }\] so the y-intercept is at the point (0, -2/3)
To get the horizontal asymptote: divide every individual block in the equation by the highest power of x, which is x^2
no x inct? no range?
o lol, forgot about range and there's no x-int as I've explained
okay
sorry, I need to a lecture now, but here is the graph for reference. good luck :D
okay.
anyone?

Not the answer you are looking for?

Search for more explanations.

Ask your own question