How do you find all the asymptotes for #sqrt(x^2-3x) - x#?
1 Answer
No vertical asymptote, Right horizontal asymptote
Explanation:
Vertical: None
Horizontal Left: None, Right:
Left:
So there is no horizontal asymptote on the left.
Right:
# = lim_(xrarroo)((sqrt(x^2-3x) - x))/1 * ((sqrt(x^2-3x) + x))/((sqrt(x^2-3x) + x)) #
# = lim_(xrarroo)(x^2-3x - x^2) /(sqrt(x^2-3x) + x) #
# = lim_(xrarroo)(-3x) /(sqrt(x^2)sqrt(1-3/x) + x) #
# = lim_(xrarroo)(-3x) /(x(sqrt(1-3/x) + 1) #
# = -3/(sqrt1+1) = -3/2#
So
Oblique Left:
There is an oblique asymptote on the left, with slope
And
There is a left oblique asymptote of the form:
We know that the vertical distance between the graph of
# = lim_(xrarr-oo)(sqrt(x^2-3x)+(x-b))# (has form#oo-oo# )
# = lim_(xrarr-oo)((x^2-3x)-(x-b)^2)/(sqrt(x^2-3x)-(x-b))#
# = lim_(xrarr-oo)((x^2-3x)-(x^2-2bx+b^2))/(sqrt(x^2)sqrt(1-3/x)-(x-b))#
# = lim_(xrarr-oo)(-3x +2bx - b^2)/(-xsqrt(1-3/x)-x-b))#
# = lim_(xrarr-oo)(-3 +2b - b^2/x)/(-sqrt(1-3/x)-1-b/x)#
# = (-3+2b)/(-1-1) = (-3+2b)/-2#
In order for this limit to be
So the oblique asymptote on the left is