Athanasios Kontogianniswww.smallmathapps.co.uk

1

Vertical asymptote x = 0 (whenever denominator equals 0)

Horizontal asymptote y = 0 (look at the behaviour of the function for large values of x)

No extreme values dy/dx ≠ 0

2

3

Vertical asymptote x = -b/a (whenever denominator equals 0)

Horizontal asymptote y = 0 (look at the behaviour of the function for large values of x)

No extreme values dy/dx ≠ 0 Cuts the y-axis at 1/b ( for x = 0) If the coefficient of x is positive then the

branches of the graph are on the 1st and 3rd quadrants.

4

5

X = 1/2 V.A.

y = 0 H.A.cuts y-axis at y = -1

6

x= -2/3 V.A. cuts y- axis at 1/2

y = 0 H.A.

If the coefficient of x is negative then the branches of the graph are on the 2nd and 4th quadrants (symmetry about the y-axis).

Alternatively if the coefficient of x is positive but there is a minus sign in front of y=f(x) (symmetry about the x-axis).

Example: y=1/(-x+1) = -1/(x-1)

7

8

Branches on 2nd and 4th quadrants since coefficient of x is negative

Stretches the function y = 1/x by a factor of a along the y-axis

Function in general has the same shape just stretched out.

9

10

y=1/x

y=3/x

vertical stretch

Draw the following graphs stating the asymptotes and where the graph cuts the axes.

y=1/(x+1) y= 2/(3x-1) y= 3/(4 – 2x) y = - 1/(x-1)

11

Check

Check

Check

Check

Next

12

x=-1 V.A.

y=0 H.A.

cuts y-axis at y=1

Back

13

x=1/3 V.A.

y=0 V.A.

cuts y-axis at y=-2

Back

14

x=2 V.A.

cuts y-axis at y=3/4

y=0 H.A.

Back

15

y=1 x=1 VA

y=0 HA

Back

Perform the Euclidean algorithm Horizontal asymptote y = k which follows

from the Euclidean algorithm The remaining part is in the form a/(bx + c)

16

3 x-2 √ 3x + 1 ∴ 3x + 1 = 3(x-2) + 7

3x – 6 ⇒ (3x + 1)/(x – 2) = 3 + 7/(x – 2 )

7

17

18

x=2 VA y=3 HA

cuts the y-axis at y = -1/2

Draw the following graphs stating the asymptotes and where the graph cuts the axes.

y= (2x – 1)/(x+1) y= (3x – 1)/(x+2) y= (3x – 1)/(x – 2)

19

Check

Check

Check

Next

20

x=-1 VA

y=2 HA

cuts y-axis at y=-1

Back

21

x = -2 VA y = 3 HA

cuts y-axis at y = -1/2

Back

22

x = 2 VA

cuts y-axis at y = 1/2

Back

y = 3 HA

For the horizontal asymptote (HA) divide numerator and denominator by x and look at large values of x i.e. if y = (3x+2)/(2x-2) =

= (3+2/x)/(2-2/x) ⇒ limy as x⇢∞ is 3/2 or in other words the ratio of the coefficients of x. Let’s look at HA

y= (2x – 1)/(4x+1) ⇒ HA = 2/4 = 1/2 y= (3x – 1)/(5x+2) ⇒ HA = 3/5 y= (3x – 1)/(1x – 2) ⇒ HA = 3

23

Let’s graph y= (3x – 1)/(5x+2) HA: y = 3/5 VA : x = -2/5 y-intercept: y = -1/2

Let’s sketch these

24

Check

Next

25

x = -2/5 VA

y = 3/5 HA

cuts y-axis at y = -1/2

Back

So y= (3x – 1)/(5x+2) has branches on the 2nd and 4th quadrants.

26

27