anonymous
  • anonymous
For the following function find the vertical, horizontal, and oblique asymptotes, if any f(x)= -23x^2/2x^2-3x^2-18x +27
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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dumbcow
  • dumbcow
is that supposed to be 2x^3 in denominator ?
anonymous
  • anonymous
yes
dumbcow
  • dumbcow
since the degree in denominator is greater, the horizontal asymptote is y=0 and there are no oblique to find the vertical asymptotes, set the denominator equal to zero and solve for x

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More answers

dumbcow
  • dumbcow
you may be able to solve it using factor by grouping
dumbcow
  • dumbcow
were you able to factor the denominator?
anonymous
  • anonymous
no
anonymous
  • anonymous
im so confused
dumbcow
  • dumbcow
2x^3-3x^2-18x +27 group them (2x^3 -3x^2) +(-18x+27) factor each group 1st group has a x^2 in common 2nd group has a -9 in common --> x^2(2x -3) -9(2x-3) Notice the terms inside parenthesis are the same, so factor them out --> (2x-3)(x^2 -9)
anonymous
  • anonymous
k
anonymous
  • anonymous
so would this be my answer
dumbcow
  • dumbcow
no you are finding the vertical asymptotes right....so your answer will be x = ? take whats above and set it equal to 0
anonymous
  • anonymous
how would i set it to equal 0
dumbcow
  • dumbcow
put a "=0" after it ??
anonymous
  • anonymous
ok i did that
dumbcow
  • dumbcow
solve for x: 2x-3 = 0 x^2 -9 = 0
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok what do i have to do now
dumbcow
  • dumbcow
solve the 2 equations...this is algebra, get x by itself if you're taking a class where you have to find asymptotes and evaluate functions, im assuming you know how to solve simple algebraic equations
anonymous
  • anonymous
x=3/2 and x=-3 or x=3
dumbcow
  • dumbcow
yep :)
anonymous
  • anonymous
so i'm done
anonymous
  • anonymous
ty

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