anonymous
  • anonymous
What is the equation of the oblique asymptote? h(x) = x^2-3x-4/x+1 a y=x+4 b y=x^2-3 c y=x d y=x-4
Mathematics
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anonymous
  • anonymous
What is the equation of the oblique asymptote? h(x) = x^2-3x-4/x+1 a y=x+4 b y=x^2-3 c y=x d y=x-4
Mathematics
schrodinger
  • schrodinger
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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.

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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
divide
misty1212
  • misty1212
you will get a quotient and a remainder ignore the remainder, the quotient is your obliques asymptote

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anonymous
  • anonymous
So the answer is D?
anonymous
  • anonymous
campbell_st
  • campbell_st
that's correct
anonymous
  • anonymous
awesome Thanks!
campbell_st
  • campbell_st
well it's not awesome as the equation doesn't have an oblique asymptote... if you factor the numerator \[\frac{x^2 - 3x -4}{x +1} = \frac{(x - 4)(x + 1)}{x +1} \] you'll find the equation simplifies to a linear form y = x - 4 and there is a point of discontinuity at x = -1 hope it helps

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