Athanasios Kontogiannis 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

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 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 1 st and 3 rd 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 2 nd and 4 th 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 2 nd and 4 th 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 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=1x=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 Next

20 x=-1 VA y=2 HA cuts y-axis at y=-1 Back

21 x = -2 VAy = 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 2 nd and 4 th quadrants. 26

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