Asymptotes of graphs of functions

Oblique Dividing numerator and denominator X2 + 3x + 1 divide by x – 1 = (x + 2) + ii) Cross multiply (x-1) ((x + 2) + ) = x2 + 3x + 1 Opening the bracket and collecting like terms together 4x = -3 X = It crosses the oblique asymptotes at x = Q2) For domain x3 + x2 > 0 x2(x + 1) > 0 x = 0 and

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}

ii) x3 + x2 = 0
x2(x + 1) = 0
x = 0 and x = -1
iii) The end behavior is y = 0 (horizontal line). Since the degree of the denominator is
larger than that of the numerator.
iv) No, there are no values that can be substituted to both the numerator and denominator
for the solution to be zero at one particular substitute.
Q3)
a) x2 + x−12
x2−4
The domain
x2 – 4 = 0
x = -2 and x = 2
domain of F(x)
{x ∈ R∨x ≠ 2 ,−2 }
b) Numerator => x2 + x – 12
x2 + 4x – 3x – 12
x(x+4)-3(x+4)
denominator
x2 – 4
x2 – 2x + 2x -4
x(x - 2)+2(x-2)
(x-2)(x+2)
F(x) = ( x−3)(x +4 )
( x−2)(x+2)
c) X – intercept are the root of the numerator
(x−3)(x +4 ) = 0
x=3∧x=−4
d) Y – intercept we get by substituting x = 0

F ( x )= x2+ x −12
x2−4
02 +0−12
02−4
Y = 3
e) since the degree of function is odd and leading coefficient is negative
so as x tends to negative infinity , f(x) tends to negative infinity
as x tends to positive infinity , f(x) tends to negative infinity
f) vertical asymptote occurs when the denominator is zero
x2 – 4 = 0
x = 2 and x = -2
Behavior of the graph
x→ -2 from left 2 from right
F(x) = (x−3)(x +4 )
( x−2)(x+2)
¿ ¿ ¿ ¿
∞ −∞
g) horizontal asymptote , y =1
h) x2 + x−12
x2−4 =1
Cross multiplying
x2+ x−12=x2−4
X = 1

Point = (8, 1)
i) The graph is above x –axis considering position of point (8, 1)
j) No, there are no values that can be substituted to both the numerator and denominator for
the solution to be zero at one particular substitute.
k)
-15 -10 -5 0 5 10 15
0
0.5
1
1.5
2
2.5
3
3.5
x
F(x)
Red line = y intercept
Black lines = x – intercepts