Asymptotes of graphs of functions

Q1)
a)
i) Vertical asymptotes
5x – 6 = 0
X = 6/5
Horizontal asymptotes
The degrees of the numerator and the denominator are equal (they are both of degree 1). The
ratio of their coefficients is 4/5. This means there will be a horizontal asymptote at y=4/5
Oblique asymptotes
Oblique asymptotes occur when the degree of denominator is lower than that of the numerator,
therefore existence of oblique is none
ii) The graph of a function cannot intersect a vertical asymptote it can meet the vertical
asymptote but cannot cross it.
iii) Numerator = 4x + 3
Presenting it in the form (x – a)k
k = degree of multiplicity
0 = -(x - -4/3)1
a = -4/3
the zero of the numerator is at -4/3, thus it is a zero of multiplicity of 1
iv) Since the multiplicity is odd then the graph will cross at -4/3
b)
i) Vertical:
x – 1 = 0
x = 1
Horizontal

In this function, there will be no horizontal asymptote, since the denominator is neither of equal
or of larger degree than the numerator.
Oblique
Dividing numerator and denominator
X2 + 3x + 1 divide by x – 1
= (x + 2) + 3
x−1
ii)
x2 +3 x +1
x−1 =(x +2)+ 3
x−1
Cross multiply
(x-1) ((x + 2) + 3
x−1) = x2 + 3x + 1
Opening the bracket and collecting like terms together
4x = -3
X = −3
4
It crosses the oblique asymptotes at x = −3
4
Q2)
i) For domain
x3 + x2 > 0
x2(x + 1) > 0
x = 0 and x = -1
domain of h(x)
{x∈ R∨x ≠ 0 ,−1}