WISKUNDE
Graad 11
NOG OEFENINGE
Hyperboliese grafieke, hiperbole : antwoorde.
  
  
Antwoorde  1
    
$$ \hspace*{2 mm}\mathrm{1.1\kern3mmy = \frac{2}{x + 1} + 3\kern2mm\ } $$

           Horisontale asimptoot : y = 3
           Vertikale asimptoot : x = − 1
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \kern5mm\ :\ \frac{2}{x + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 1)\ \ :\ \kern3mm\ 2 + 3(x + 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{2 + 3x + 3 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{5}{3} = −1,67\kern2mm\ } $$

             X-afsnit is (−1,67 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{2}{0 + 1} + 3\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= 5\kern2mm\ } $$
             Y-afsnit is (0 ; 5)
             a is positief en dus is die grafiek
             in kwadrante I en III
  

                                                                   [ V 1.1 ]
    
$$ \hspace*{2 mm}\mathrm{1.2\kern3mmy = \frac{3}{x + 2} − 4\kern2mm\ } $$

           Horisontale asimptoot : y = −4
           Vertikale asimptoot : x = − 2
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \kern5mm\ :\ \frac{3}{x + 2} − 4 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 2)\ \ :\ \kern3mm\ 3 − 4(x + 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 − 4x − 8 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{5}{4} = −1,25\kern2mm\ } $$

             X-afsnit is (−1,25 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{3}{0 + 2} − 4\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= −\frac{5}{2} = − 2,5\kern2mm\ } $$

             Y-afsnit is (0 ; −2,5)
             a is positief en dus is die grafiek in
             kwadrante I en III
  

                                                                   [ V 1.2 ]
    
$$ \hspace*{2 mm}\mathrm{1.3\kern3mmy = \frac{5}{x − 3} + 5\kern2mm\ } $$

           Horisontale asimptoot : y = 5
           Vertikale asimptoot : x = 3
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \kern5mm\ :\ \frac{5}{x − 3} + 5 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x − 3)\ \ :\ \kern3mm\ 5 + 5(x − 3) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{5 + 5x − 15 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{10}{5} = 2\kern2mm\ } $$

             X-afsnit is (2 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{5}{0 − 3} + 5\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= \frac{10}{3} = 3,33\kern2mm\ } $$

             Y-afsnit is (0 ; 3,33)
             a is positief en dus is die grafiek in
             kwadrante I en III
  

                                                                   [ V 1.3 ]
    
$$ \hspace*{2 mm}\mathrm{1.4\kern3mmy = \frac{−2}{x + 1} + 3\kern2mm\ } $$

           Horisontale asimptoot : y = 3
           Vertikale asimptoot : x = − 1
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern6mm\ \frac{−2}{x + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 1)\ \ :\ \kern3mm\ −2 + 3(x + 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 2 + 3x + 3 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{1}{3} = − 0,33\kern2mm\ } $$

             X-afsnit is (− 0,33 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{−2}{0 + 1} + 3\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= 1\kern2mm\ } $$

             Y-afsnit is (0 ; 1)
             a is negatief en dus is die grafiek in
             kwadrante II en IV
  

                                                                   [ V 1.4 ]
    
$$ \hspace*{2 mm}\mathrm{1.5\kern3mmy = \frac{−3}{x − 2} − 4\kern2mm\ } $$

           Horisontale asimptoot : y = − 4
           Vertikale asimptoot : x = 2
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern6mm\ \frac{−3}{x − 2} − 4 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x− 2)\ \ :\ \kern3mm\ −3 − 4(x − 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 3 − 4x + 8 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{5}{4} = 1,25\kern2mm\ } $$

             X-afsnit is (1,25 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{−3}{0 − 2} − 4\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= −\frac{5}{2} \ =\ \ −2,5\kern2mm\ } $$

             Y-afsnit is (0 ; − 2,5)
             a is negatief en dus is die grafiek in
             kwadrante II en IV
  

                                                                   [ V 1.5 ]
    
$$ \hspace*{2 mm}\mathrm{1.6\kern3mmy = \frac{−5}{x + 3} − 2\kern2mm\ } $$

           Horisontale asimptoot : y = − 2
           Vertikale asimptoot : x = − 3
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern6mm\ \frac{−5}{x + 3} − 2 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 3)\ \ :\ \kern3mm\ −5 − 2(x + 3) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 5 − 2x − 6 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{11}{2} = −5,5\kern2mm\ } $$

             X-afsnit is (−5,5 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{−5}{0 + 3} − 2\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= −\frac{11}{3} \ =\ \ −3,67\kern2mm\ } $$

             Y-afsnit is (0 ; − −3,67)
             a is negatief en dus is die grafiek in
             kwadrante II en IV
  

                                                                   [ V 1.6 ]
  
Antwoorde  2
    
     2.1  Horisontale asimptoot: y = 3 en
            dus q = 3
            Vertic\kale asimptoot : x = − 1 en
            dus p = 1
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x + 1} + 3\kern2mm\ } $$

           Vervang nou die koördinate van 'n punt
           vir x en y in die vergelyking en
           los op vir a.
$$ \hspace*{9 mm}\mathrm{By\ A(−2;0)\ :\ \frac{a}{− 2 + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{− 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X − 1\ \ :\ a − 3 = 0\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{3}{x + 1} + 3\kern2mm\ } $$

                                                                 [ V 2.1 ]
    
     2.2  Horisontale asimptoot: y = − 3 en
            dus q = − 3
            Vertikale asimptoot : x = 1 en
            dus p = − 1
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x − 1} − 3\kern2mm\ } $$

           Vervang nou die koördinate van 'n punt
           vir x en y in die vergelyking en
           los op vir a.
$$ \hspace*{9 mm}\mathrm{By\ B(0;8)\ :\ \frac{a}{0 − 1} − 3 = 8\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{− 1} − 3 = 8\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X − 1\ \ :\ a + 3 = 8\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{a = 5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{5}{x − 1} − 3\kern2mm\ } $$

                                                                 [ V 2.2 ]
    
     2.3  P is die punt (− 2; 4) en dus
            horisontale asimptoot: y = 4 en
            dus q = 4
            vertikale asimptoot : x = − 2 en
            dus p = 2
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x + 2} + 4\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{By\ Q(−3;7)\ :\ \frac{a}{−3 + 2} + 4 = 7\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{− 1} + 4 = 7\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X − 1\ \ :\ a − 4 = − 7\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{a = − 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{− 3}{x + 2} + 4\kern2mm\ } $$

                                                                 [ V 2.3 ]
    
     2.4  P is die punt (3 ; -1) en dus
            horisontale asimptoot: y = -1 en
            dus q = − 1
            vertikale asimptoot : x = 3 en
            dus p = -3
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x - 3} − 1\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{By\ Q(5 ; −3)\ :\ \frac{a}{5 − 3} − 1 = − 3\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{2} − 1 = − 3\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X 2\ \ :\ a − 2 = − 6\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = − 4\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{− 4}{x − 3} − 1\kern2mm\ } $$

                                                                 [ V 2.4 ]
  
Antwoorde  3
    
     3.1  P is die punt (1 ; 2)                      [ V 3.1 ]
    
     3.2  Horisontale asimptoot: y = 2 en
            dus q = 2
            Vertikale asimptoot : x = 1 en
            dus p = -1
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x - 1} + 2\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{By\ Q(2 ; 5)\ :\ \frac{a}{2 − 1} + 2 = 5\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{\frac{a}{1} + 2 = 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X 1\ \ :\ a + 2 = 5\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{3}{x − 1} + 2\kern2mm\ } $$
                                                                     [ V 3.2 ]
    
$$ \hspace*{2 mm}\mathrm{3.3\kern3mmVergelyking\ is\ y = \frac{3}{x − 1} + 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern4mm\ \frac{3}{x − 1} + 2 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x − 1)\ \ :\ \kern3mm\ 3 + 2(x − 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 + 2x − 2 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{−1}{2} = − 0,5\kern2mm\ } $$

             X-afsnit, punt A is (− 0,5 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{3}{0 − 1} + 2\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= −1\kern2mm\ } $$

             Y-afsnit, punt B is (0 ; − 1)
                                                                     [ V 3.3 ]
    
     3.4  y = (x + p) + q
            y = (x − 1) + 2
            y = x + 1                                          [ V 3.4 ]
    
     3.5  By R is die vergelykings gelyk, sodat
$$ \hspace*{34 mm}\mathrm{\frac{3}{x − 1} + 2 = x + 1\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x − 1)\ \ :\ \kern3mm\ 3 + 2(x − 1) = (x + 1)(x − 1)\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 + 2x − 2 = x^2 − 1\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{x^2 − 2x − 2 = 0\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{x = \frac{−(−2) ± \sqrt{(−2)^2 − 4(1)(−2)}}{2(1)}\kern2mm\ } $$

$$ \hspace*{31 mm}\mathrm{= \frac{2 ± \sqrt{12}}{2}\kern2mm\ } $$

$$ \hspace*{29 mm}\mathrm{x = 2,73\ of\ X = − 0,73\kern2mm\ } $$
            By R x > 0 en dus x = 2,73 en
            y = 2,73 + 1 = 3,73
            R is die punt (2,73 ; 3,73)             [ V 3.5 ]

  
Antwoorde  4
    
     4.1  P is die punt (− 2 ; − 3)                      [ V 4.1 ]
    
     4.2  Horisontale asimptoot: y = − 3 en
            dus q = − 3
            Vertikale asimptoot : x = − 2 en
            dus p = 2
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x + 2} − 3\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{By\ Q(− 1 ; 2)\ :\ \frac{a}{− 1 + 2} − 3 = 2\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{\frac{a}{1} − 3 = 2\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X 1\ \ :\ a − 3 = 2\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = 5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{5}{x + 2} − 3\kern2mm\ } $$
                                                                     [ V 4.2 ]
    
$$ \hspace*{2 mm}\mathrm{4.3\kern3mmVergelyking\ is\ y = \frac{5}{x + 2} − 3\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern4mm\ \frac{5}{x + 2} − 3 = 0\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x + 2)\ \ :\ \kern3mm\ 5 − 3(x + 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{5 − 3x − 6 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{−1}{3} = − 0,33\kern2mm\ } $$

             X-afsnit, punt A is (− 0,33 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{5}{0 + 2} − 3\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= − 0,5\kern2mm\ } $$

             Y-afsnit, punt B is (0 ; − 0,5)
                                                                     [ V 4.3 ]
    
     4.4  y = (x + p) + q
            y = (x + 2) − 3
            y = x − 1                                          [ V 4.4 ]
    
     4.5  By T is die vergelykings gelyk, sodat
$$ \hspace*{34 mm}\mathrm{\frac{5}{x + 2} − 3 = x − 1\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x + 2)\ \ :\ \kern3mm\ 5 − 3(x + 2) = (x + 2)(x − 1)\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{5 − 3x − 6 = x^2 + x − 2\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{x^2 + 4x − 1 = 0\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{x = \frac{−(4) ± \sqrt{(4)^2 − 4(1)(−1)}}{2(1)}\kern2mm\ } $$

$$ \hspace*{32 mm}\mathrm{= \frac{−4 ± \sqrt{20}}{2}\kern2mm\ } $$

$$ \hspace*{29 mm}\mathrm{x = 0,24\ of\ X = − 4,24\kern2mm\ } $$
            By T x < 0 en dus x = − 4,24 en
            y = − 4,24 − 1 = − 5,24
            T is die punt (− 4,24 ; − 5,24)             [ V 4.5 ]

    
     4.6  Definisie-versameling :
                                                     x = {x | x ≠ − 2; x ∈ ℜ}
            Waarde-versameling : y = {y | y ≠ − 3; y ∈ ℜ}
                                                                         [ V 4.6 ]

  
Antwoorde  5
    
     5.1  y = 4                                                  [ V 5.1 ]
    
     5.2  x = 2                                                  [ V 5.2 ]
    
     5.3  Horisontale asimptoot: y = 4 en
            dus q = 4
            Vertikale asimptoot : x = 2 en
            dus p = -2
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{a}{x - 2} + 4\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{By\ Q(−2 ; 6)\ :\ \frac{a}{−2 − 2} + 4 = 6\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{−4} + 4 = 6\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{X −4\ \ :\ a − 16 = − 24\kern2mm\ } $$
$$ \hspace*{45 mm}\mathrm{a = − 8\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Vergelyking\ is\ y = \frac{−8}{x − 2} + 4\kern2mm\ } $$
                                                                      [ V 5.3 ]
    
$$ \hspace*{2 mm}\mathrm{5.4\kern3mmVergelyking\ is\ y = \frac{−8}{x − 2} + 4\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern6mm] \frac{−8}{x − 2} + 4 = 0\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x − 2)\ :\ \kern3mm\ −8 + 4(x − 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 8 + 4x − 8 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{16}{4} = 4\kern2mm\ } $$

             X-afsnit, punt A is (4 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{−8}{0 − 2}+ 4\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= 8\kern2mm\ } $$

             Y-afsnit, punt B is (0 ; 8)
                                                                     [ V 5.4 ]
    
     5.5   y = − (x + p) + q
             y = − (x − 2) + 4
                = − x + 2 + 4
                = − x + 6                                       [ V 5.5 ]
    
     5.6  By R en S is die vergelykings gelyk,
            sodat
$$ \hspace*{34 mm}\mathrm{\frac{−8}{x − 2} + 4 = − x + 6\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x − 2)\ \ :\ \kern3mm\ − 8 + 4(x − 2) = (x − 2)(−x + 6)\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{−8 + 4x − 8 = −x^2 + 8x − 12\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{x^2 − 4x − 4 = 0\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{x = \frac{−(−4) ± \sqrt{(−4)^2 − 4(1)(−4)}}{2(1)}\kern2mm\ } $$

$$ \hspace*{32 mm}\mathrm{x = \frac{4 ± \sqrt{32}}{2}\kern2mm\ } $$

$$ \hspace*{29 mm}\mathrm{x = −0,83\ \ of\ \ x = 4,83\kern2mm\ } $$
                                   y = 6,83      of   y = 1,17
                                                                      [ V 5.6 ]

    
     5.7  Definisie-versameling :
                                                     x = {x | x ≠ 2; x ∈ ℜ}
            Waarde-versameling : y = {y | y ≠ 4; y ∈ ℜ}
                                                                         [ V 5.7 ]

    
$$ \hspace*{2 mm}\mathrm{5.8\kern6mmy = \frac{−8}{x − 2} + 4\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{h(x) = \frac{−8}{(x + 2) − 5} + (4 − 3)\kern2mm\ } $$

$$ \hspace*{16 mm}\mathrm{= \frac{−8}{x − 3} + 1\kern2mm\ } $$                          [ V 5.8 ]

  
Antwoorde  6
    
     6.1  Horisontale asimptoot : y = 5 en
             dus q = 5                                        [ V 6.1 ]
    
     6.2  Vertikale asimptoot : x = 3 en
             dus p = − 3                                      [ V 6.2 ]
    
$$ \hspace*{2 mm}\mathrm{6.3\kern3mmVergelyking\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern7mm\ − \frac{6}{x - 3} = 0\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x − 3)\ :\ \kern3mm\ 5(x − 3) − 6 = 0\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{5x − 15 − 6 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{21}{5} = 4,20\kern2mm\ } $$

             X-afsnit, punt A is (4,20 ; 0)
                                                                     [ V 6.3 ]
    
$$ \hspace*{2 mm}\mathrm{6.4\kern3mmVergelyking\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = 5 − \frac{6}{0 - 3}\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= 7\kern2mm\ } $$
             Y-afsnit, punt B is (0 ; 7)
                                                                     [ V 6.4 ]
    
$$ \hspace*{2 mm}\mathrm{6.5\kern3mmVergelyking\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{By\ D(5;d)\ :\ d = 5 − \frac{6}{5 - 3}\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{= 2\kern2mm\ } $$
                                                                     [ V 6.5 ]
    
$$ \hspace*{2 mm}\mathrm{6.6\kern3mmVergelyking\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{By\ E(e;6,5)\ :\ 5 − \frac{6}{e - 3} = 6,5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (e − 3)\ :\ \kern3mm\ 5(e − 3) − 6 = 6,5(e − 3)\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{5e − 15 − 6 = 6,5e\ − 19,5\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{e = \frac{−1,5}{1,5} = − 1\kern2mm\ } $$

                                                                     [ V 6.6 ]
    
     6.7  By R en S is die vergelykings gelyk,
            sodat
$$ \hspace*{27 mm}\mathrm{5 − \frac{6}{x − 3} = − 2x + 15\kern2mm\ } $$

$$ \hspace*{6 mm}\mathrm{X (x − 3)\ \ :\ 5(x − 3) − 6 = (− 2x + 15)(x − 3)\kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{5x − 15 − 6 = −2x^2 + 21x − 45\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{x^2 − 8x + 12 = 0\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{(x − 2)(x − 6) = 0\kern2mm\ } $$
                                        x = 2  of  x = 6
                                      y = 11  of  y = 3
            Die snypunte is (2 ; 11) en (6 ; 3)
                                                                      [ V 6.7 ]

  
Antwoorde  7
    
     7.1  P is die punt (− 1 ; − 5)                   [ V 7.1 ]
    
$$ \hspace*{2 mm}\mathrm{7.2\kern3mmVergelyking\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X-afsnit\ \ :\ \kern5mm\ \frac{−5}{x + 1} − 3 = 0\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X (x + 1)\ :\ \kern3mm\ − 5 − 3(x + 1) = 0\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{−5 − 3x − 3 = 0\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{x = \frac{− 8}{3} = − 2,67\kern2mm\ } $$

             X-afsnit, punt A is (− 2,67 ; 0)
                                                                     [ V 7.2 ]
    
$$ \hspace*{2 mm}\mathrm{7.3\kern3mmVergelyking\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{Y-afsnit\ \ :\ y = \frac{−5}{0 + 1} − 3\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{= − 8\kern2mm\ } $$
             Y-afsnit, punt B is (0 ; − 8)
                                                                     [ V 7.3 ]
    
$$ \hspace*{2 mm}\mathrm{7.4\kern3mmVergelyking\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{By\ D(0,5;d)\ :\ d = \frac{−5}{0,5 + 1} − 3\kern2mm\ } $$

$$ \hspace*{36 mm}\mathrm{= − 6,33\kern2mm\ } $$        [ V 7.4 ]
    
$$ \hspace*{2 mm}\mathrm{7.5\kern3mmVergelyking\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{By\ E(e;−1)\ :\ \kern2mm\ \frac{−5}{e + 1} − 3 = − 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (e + 1)\ :\ \kern3mm\ − 5 − 3(e + 1) = −(e + 1)\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{−5 − 3e − 3 = − e\ − 1\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{e = \frac{−7}{2} = − 3,5\kern2mm\ } $$

                                                                     [ V 7.5 ]
    
     7.6  Positiewe simmetrie-as : y = (x + p) + q
                                                             = (x + 1) − 5
                                                             = x − 4
             Negatiewe simmetrie-as : y = -(x + p) + q
                                                               = −(x + 1) − 5
                                                               = − x − 6
                                                                      [ V 7.6 ]
  
Antwoorde  8
    
$$ \hspace*{2 mm}\mathrm{8.1\kern3mmVergelyking\ is\ y = \frac{4}{x − 1} + 2\kern2mm\ } $$

            Horisontale asimptoot : y = 2
            Vertikale asimptoot : x = 1
            Die stukke wat pas is p en r
$$ \hspace*{10 mm}\mathrm{Vergelyking\ is\ y = \frac{4}{x + 2} − 2\kern2mm\ } $$

            Horisontale asimptoot : y = − 2
            Vertikale asimptoot : x = − 2
            Die stukke wat pas is q en s
                                                                      [ V 8.1 ]
    
$$ \hspace*{2 mm}\mathrm{8.2\kern3mmVergelyking\ is\ y = \frac{4}{x − 1} + 2 = \frac{4}{x − 1 + 3} + ( − 4)\kern2mm\ } $$

$$ \hspace*{34 mm}\mathrm{Grafiek A\ \ = \frac{4}{x + 2} − 2\kern2mm\ } $$
$$ \hspace*{53 mm}\mathrm{Grafiek B\kern2mm\ } $$

            Grafiek A word 3 eenhede na links en
            4 eenhede afwaarts transleer om
            grafiek B te vorm.
                                                                      [ V 8.2 ]