UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1613: Calculus B
Assignment 1 — Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1613
Module Code:
Calculus B
Module Name:
Assignment 1
Assignment:
15 May 2026
Due Date:
67
Total Marks:
Submitted in partial fulfilment of the requirements for Calculus B — UNISA 2026
,UNISA | MAT1613 Calculus B – Assignment 1
Question 1: Asymptotes, Curve Sketching and Extreme Points [21 Marks]
√
2x2 + 1
Question 1.1 — Asymptotes of f (x) = [5 Marks]
3x − 5
Find the horizontal and vertical asymptote of
√
2x2 + 1
f (x) =
3x − 5
Step 1: Vertical Asymptote
The vertical asymptote occurs where the denominator equals zero.
Set the denominator equal to zero:
5
3x − 5 = 0 =⇒ x =
3
5
Vertical asymptote: x =
3
Step 2: Horizontal Asymptote as x → +∞
Start with the original function: √
2x2 + 1
f (x) =
3x − 5
Factor x2 out of the square root in the numerator:
s
1
x2 2+ 2
x
=
3x − 5
√
Since x2 = x for x > 0: r
1
x 2+
x2
=
3x − 5
Page 2 of 16
, UNISA | MAT1613 Calculus B – Assignment 1
Divide every term in numerator and denominator by x:
r
1
2+
x2
=
5
3−
x
1 5
As x → +∞, both 2
→ 0 and → 0, therefore:
x x
√ √
2+0 2
lim f (x) = =
x→+∞ 3−0 3
Step 3: Horizontal Asymptote as x → −∞
√ q
1
For x < 0, x2 = |x| = −x, so the numerator becomes −x 2+ x2
.
Dividing numerator and denominator by x (negative):
r
1 √
− 2+
x2 − 2
= →
5 3
3−
x
Key Distinction
√
This function has two distinct horizontal asymptotes because x2 evaluates differ-
ently depending on the sign of x. Many students overlook the x → −∞ case, which
yields a different value.
Horizontal asymptotes:
√ √
2 2
y= as x → +∞ and y=− as x → −∞
3 3
Page 3 of 16
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1613: Calculus B
Assignment 1 — Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1613
Module Code:
Calculus B
Module Name:
Assignment 1
Assignment:
15 May 2026
Due Date:
67
Total Marks:
Submitted in partial fulfilment of the requirements for Calculus B — UNISA 2026
,UNISA | MAT1613 Calculus B – Assignment 1
Question 1: Asymptotes, Curve Sketching and Extreme Points [21 Marks]
√
2x2 + 1
Question 1.1 — Asymptotes of f (x) = [5 Marks]
3x − 5
Find the horizontal and vertical asymptote of
√
2x2 + 1
f (x) =
3x − 5
Step 1: Vertical Asymptote
The vertical asymptote occurs where the denominator equals zero.
Set the denominator equal to zero:
5
3x − 5 = 0 =⇒ x =
3
5
Vertical asymptote: x =
3
Step 2: Horizontal Asymptote as x → +∞
Start with the original function: √
2x2 + 1
f (x) =
3x − 5
Factor x2 out of the square root in the numerator:
s
1
x2 2+ 2
x
=
3x − 5
√
Since x2 = x for x > 0: r
1
x 2+
x2
=
3x − 5
Page 2 of 16
, UNISA | MAT1613 Calculus B – Assignment 1
Divide every term in numerator and denominator by x:
r
1
2+
x2
=
5
3−
x
1 5
As x → +∞, both 2
→ 0 and → 0, therefore:
x x
√ √
2+0 2
lim f (x) = =
x→+∞ 3−0 3
Step 3: Horizontal Asymptote as x → −∞
√ q
1
For x < 0, x2 = |x| = −x, so the numerator becomes −x 2+ x2
.
Dividing numerator and denominator by x (negative):
r
1 √
− 2+
x2 − 2
= →
5 3
3−
x
Key Distinction
√
This function has two distinct horizontal asymptotes because x2 evaluates differ-
ently depending on the sign of x. Many students overlook the x → −∞ case, which
yields a different value.
Horizontal asymptotes:
√ √
2 2
y= as x → +∞ and y=− as x → −∞
3 3
Page 3 of 16
