How do you find the vertical, horizontal and slant asymptotes of: #h(x)= (x^2-4)/(x)#?
1 Answer
vertical asymptote at x = 0
slant asymptote y = x
Explanation:
The denominator of h(x) cannot be zero as this would make h(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve:
#x=0rArrx=0" is the asymptote"#
#color(blue)"Horizontal asymptotes"# occur when the degree of the numerator ≤ degree of the denominator. This is not the case here (numerator degree 2 , denominator degree 1 ) Hence there are no horizontal asymptotes.
#color(blue)"slant asymptotes"# occur when the degree of the numerator > degree of denominator. Hence there is a slant asymptote.divide numerator by denominator.
#rArrh(x)=(x^2)/x-4/x=x-4/x# as
#xto+-oo,h(x)tox-0#
#rArry=x" is the asymptote"#
graph{(x^2-4)/x [-10, 10, -5, 5]}