Asked by changaya
for the curve y=(x^2+1)/(x^2-4), find:
(i). The cordinates of the turning points
(ii). The equations of the asymptotes
(iii). Sketch the curve
(i). The cordinates of the turning points
(ii). The equations of the asymptotes
(iii). Sketch the curve
Answers
Answered by
Reiny
dy/dx = (2x(x^2 -4) - 2x(x^2+1))/(x^2-4)^2
= (2x^3 - 8x - 2x^3 - 2x)/(x^2-4)^2
= -6x/(x^2-4)^2
= 0 for max/min or turning points
6x/(x^2-4)^2 = 0 ---> x = 0
then y = -1/4
the turning point is (0, -1/4)
asymtotes .... when the denominator is zero
x^2 - 4 = 0
x = ± 2
VA at x = 2 and x = -2
http://www.wolframalpha.com/input/?i=plot+y%3D%28x%5E2%2B1%29%2F%28x%5E2-4%29
= (2x^3 - 8x - 2x^3 - 2x)/(x^2-4)^2
= -6x/(x^2-4)^2
= 0 for max/min or turning points
6x/(x^2-4)^2 = 0 ---> x = 0
then y = -1/4
the turning point is (0, -1/4)
asymtotes .... when the denominator is zero
x^2 - 4 = 0
x = ± 2
VA at x = 2 and x = -2
http://www.wolframalpha.com/input/?i=plot+y%3D%28x%5E2%2B1%29%2F%28x%5E2-4%29
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