anonymous
  • anonymous
I have a question on how to sketch a graph...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
i figured out the derivatives but how do i get the x-int y-int?
AccessDenied
  • AccessDenied
X-intercepts occur when y=0 (touches the x-axis) Y-intercepts occur when x=0 (touches the y-axis)

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apoorvk
  • apoorvk
put x=0, and get y for for the value of y-int. viceversa for x-int.
anonymous
  • anonymous
for the numerator or denominator?
anonymous
  • anonymous
what about horizontal asymptote?
AccessDenied
  • AccessDenied
This refers to the y and x in y = 1/(x-2) - 3 Try evaluating for when y=0 and when x=0 and solve for the other variable. Those will be the intercept points (x,0) and (0,y) (respectively.)
AccessDenied
  • AccessDenied
We could either use the fact that 1/(x-2) is a shift to the right of 1/x and that we know 1/x has horizontal asymptote y=0... or we could look at the function as a quotient of two polynomials and only look at the leading terms and their degrees. when the degree of the numerator is less than or equal to the degree of the denominator, there is a horizontal asymptote. * equal => (numerator leading coefficient)/(denominator leading coefficient) * less than => y=0

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