MATHEMATICS
Grade 11
MORE EXERCISES
Hyperbolic graphs, hyperboles : answers.
  
  
Answers  1
    
$$ \hspace*{2 mm}\mathrm{1.1\kern3mmy = \frac{2}{x + 1} + 3\kern2mm\ } $$

           Horizontal asymptote : y = 3
           Vertical asymptote : x = − 1
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{2}{x + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 1)\ \ :\ \kern3mm\ 2 + 3(x + 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{2 + 3x + 3 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{5}{3} = −1,67\kern2mm\ } $$

             X-intercept is (−1,67 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{2}{0 + 1} + 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 5\kern2mm\ } $$
             Y-intercept is (0 ; 5)
             a is positive and thus the graph is
             in quadrants I and III
  

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

           Horizontal asymptote : y = −4
           Vertical asymptote : x = − 2
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{3}{x + 2} − 4 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 2)\ \ :\ \kern3mm\ 3 − 4(x + 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 − 4x − 8 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{5}{4} = −1,25\kern2mm\ } $$

             X-intercept is (−1,25 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{3}{0 + 2} − 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −\frac{5}{2} = − 2,5\kern2mm\ } $$

             Y-intercept is (0 ; −2,5)
             a is positive and thus the graph is
             in quadrants I and III
  

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

           Horizontal asymptote : y = 5
           Vertical asymptote : x = 3
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{5}{x − 3} + 5 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x − 3)\ \ :\ \kern3mm\ 5 + 5(x − 3) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{5 + 5x − 15 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{10}{5} = 2\kern2mm\ } $$

             X-intercept is (2 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{5}{0 − 3} + 5\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= \frac{10}{3} = 3,33\kern2mm\ } $$

             Y-intercept is (0 ; 3,33)
             a is positive and thus the graph is
             in quadrants I and III
  

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

           Horizontal asymptote : y = 3
           Vertical asymptote : x = − 1
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{−2}{x + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 1)\ \ :\ \kern3mm\ −2 + 3(x + 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 2 + 3x + 3 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{1}{3} = − 0,33\kern2mm\ } $$

             X-intercept is (− 0,33 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{−2}{0 + 1} + 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 1\kern2mm\ } $$

             Y-intercept is (0 ; 1)
             a is negative and thus the graph is
             in quadrants II and IV
  

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

           Horizontal asymptote : y = − 4
           Vertical asymptote : x = 2
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{−3}{x − 2} − 4 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x− 2)\ \ :\ \kern3mm\ −3 − 4(x − 2) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 3 − 4x + 8 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{5}{4} = 1,25\kern2mm\ } $$

             X-intercept is (1,25 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{−3}{0 − 2} − 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −\frac{5}{2} \ =\ \ −2,5\kern2mm\ } $$

             Y-intercept is (0 ; − 2,5)
             a is negative and thus the graph is
             in quadrants II and IV
  

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

           Horizontal asymptote : y = − 2
           Vertical asymptote : x = − 3
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{−5}{x + 3} − 2 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x + 3)\ \ :\ \kern3mm\ −5 − 2(x + 3) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{− 5 − 2x − 6 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = −\frac{11}{2} = −5,5\kern2mm\ } $$

             X-intercept is (−5,5 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{−5}{0 + 3} − 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −\frac{11}{3} \ =\ \ −3,67\kern2mm\ } $$

             Y-intercept is (0 ; − −3,67)
             a is negative and thus the graph is
             in quadrants II and IV
  

                                                                   [ Q 1.6 ]
  
Answers  2
    
     2.1  Horizontal asymptote: y = 3 and
            thus q = 3
            Vertical asymptote : x = − 1 and
            thus p = 1
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x + 1} + 3\kern2mm\ } $$

           Now substitute the coordinates of
           a point for x and y in the equation
           and solve for a.
$$ \hspace*{9 mm}\mathrm{At\ A(−2;0)\ :\ \frac{a}{− 2 + 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{− 1} + 3 = 0\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X − 1\ \ :\ a − 3 = 0\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{3}{x + 1} + 3\kern2mm\ } $$

                                                                 [ Q 2.1 ]
    
     2.2  Horizontal asymptote: y = − 3 and
            thus q = − 3
            Vertical asymptote : x = 1 and
            thus p = − 1
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x − 1} − 3\kern2mm\ } $$

           Now substitute the coordinates of
           a point for x and y in the equation
           and solve for a.
$$ \hspace*{9 mm}\mathrm{At\ B(0;8)\ :\ \frac{a}{0 − 1} − 3 = 8\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{− 1} − 3 = 8\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X − 1\ \ :\ a + 3 = 8\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{a = 5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{5}{x − 1} − 3\kern2mm\ } $$

                                                                 [ Q 2.2 ]
    
     2.3  P is the point (− 2; 4) and thus
            horizontal asymptote: y = 4 and
            thus q = 4
            vertical asymptote : x = − 2 and
            thus p = 2
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x + 2} + 4\kern2mm\ } $$

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

                                                                 [ Q 2.3 ]
    
     2.4  P is the point (3 ; -1) and thus
            horizontal asymptote: y = -1 and
            thus q = − 1
            vertical asymptote : x = 3 and
            thus p = -3
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x - 3} − 1\kern2mm\ } $$

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

                                                                 [ Q 2.4 ]
  
Answers  3
    
     3.1  P is the point (1 ; 2)                      [ Q 3.1 ]
    
     3.2  Horizontal asymptote: y = 2 and
            thus q = 2
            Vertical asymptote : x = 1 and
            thus p = -1
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x - 1} + 2\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{At\ Q(2 ; 5)\ :\ \frac{a}{2 − 1} + 2 = 5\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{\frac{a}{1} + 2 = 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X 1\ \ :\ a + 2 = 5\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{3}{x − 1} + 2\kern2mm\ } $$
                                                                     [ Q 3.2 ]
    
$$ \hspace*{2 mm}\mathrm{3.3\kern3mmEquation\ is\ y = \frac{3}{x − 1} + 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \frac{3}{x − 1} + 2 = 0\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{X (x − 1)\ \ :\ \kern3mm\ 3 + 2(x − 1) = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 + 2x − 2 = 0\kern2mm\ } $$
$$ \hspace*{39 mm}\mathrm{x = \frac{−1}{2} = − 0,5\kern2mm\ } $$

             X-intercept, point A is (− 0,5 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{3}{0 − 1} + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −1\kern2mm\ } $$

             Y-intercept, point B is (0 ; − 1)
                                                                     [ Q 3.3 ]
    
     3.4  y = (x + p) + q
            y = (x − 1) + 2
            y = x + 1                                          [ Q 3.4 ]
    
     3.5  At R the equations are equal, so that
$$ \hspace*{34 mm}\mathrm{\frac{3}{x − 1} + 2 = x + 1\kern2mm\ } $$

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

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

$$ \hspace*{29 mm}\mathrm{x = 2,73\ or\ X = − 0,73\kern2mm\ } $$
            At R x > 0 and thus x = 2,73 and
            y = 2,73 + 1 = 3,73
            R is the point (2,73 ; 3,73)             [ Q 3.4 ]

  
Answers  4
    
     4.1  P is the point (− 2 ; − 3)                      [ Q 4.1 ]
    
     4.2  Horizontal asymptote: y = − 3 and
            thus q = − 3
            Vertical asymptote : x = − 2 and
            thus p = 2
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x + 2} − 3\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{At\ Q(− 1 ; 2)\ :\ \frac{a}{− 1 + 2} − 3 = 2\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{\frac{a}{1} − 3 = 2\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{X 1\ \ :\ a − 3 = 2\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{a = 5\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{5}{x + 2} − 3\kern2mm\ } $$
                                                                     [ Q 4.2 ]
    
$$ \hspace*{2 mm}\mathrm{4.3\kern3mmEquation\ is\ y = \frac{5}{x + 2} − 3\kern2mm\ } $$

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

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

             X-intercept, point A is (− 0,33 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{5}{0 + 2} − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= − 0,5\kern2mm\ } $$

             Y-intercept, point B is (0 ; − 0,5)
                                                                     [ Q 4.3 ]
    
     4.4  y = (x + p) + q
            y = (x + 2) − 3
            y = x − 1                                          [ Q 4.4 ]
    
     4.5  At T the equations are equal, so that
$$ \hspace*{34 mm}\mathrm{\frac{5}{x + 2} − 3 = x − 1\kern2mm\ } $$

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

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

$$ \hspace*{29 mm}\mathrm{x = 0,24\ or\ X = − 4,24\kern2mm\ } $$
            At T x < 0 and thus x = − 4,24 and
            y = − 4,24 − 1 = − 5,24
            T is the point (− 4,24 ; − 5,24)             [ Q 4.5 ]

    
     4.6  Domain : x = {x | x ≠ − 2; x ∈ ℜ}
            Range : y = {y | y ≠ − 3; y ∈ ℜ}
                                                                         [ Q 4.6 ]

  
Answers  5
    
     5.1  y = 4                                                  [ Q 5.1 ]
    
     5.2  x = 2                                                  [ Q 5.2 ]
    
     5.3  Horizontal asymptote: y = 4 and
            thus q = 4
            Vertical asymptote : x = 2 and
            thus p = -2
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{a}{x - 2} + 4\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{At\ Q(−2 ; 6)\ :\ \frac{a}{−2 − 2} + 4 = 6\kern2mm\ } $$
$$ \hspace*{36 mm}\mathrm{\frac{a}{−4} + 4 = 6\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{X −4\ \ :\ a − 16 = − 24\kern2mm\ } $$
$$ \hspace*{45 mm}\mathrm{a = − 8\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ is\ y = \frac{−8}{x − 2} + 4\kern2mm\ } $$
                                                                      [ Q 5.3 ]
    
$$ \hspace*{2 mm}\mathrm{5.4\kern3mmEquation\ is\ y = \frac{−8}{x − 2} + 4\kern2mm\ } $$

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

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

             X-intercept, point A is (4 ; 0)
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{−8}{0 − 2}+ 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 8\kern2mm\ } $$

             Y-intercept, point B is (0 ; 8)
                                                                     [ Q 5.4 ]
    
     5.5   y = − (x + p) + q
             y = − (x − 2) + 4
                = − x + 2 + 4
                = − x + 6                                       [ Q 5.5 ]
    
     5.6  At R and S the equations are equal,
            so that
$$ \hspace*{34 mm}\mathrm{\frac{−8}{x − 2} + 4 = − x + 6\kern2mm\ } $$

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

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

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

    
     5.7  Domain : x = {x | x ≠ 2; x ∈ ℜ}
            Range : y = {y | y ≠ 4; y ∈ ℜ}
                                                                         [ Q 5.7 ]

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

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

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

  
Answers  6
    
     6.1  Horizontal asymptote : y = 5 and
             thus q = 5                                        [ Q 6.1 ]
    
     6.2  Vertical asymptote : x = 3 and
             thus p = − 3                                      [ Q 6.2 ]
    
$$ \hspace*{2 mm}\mathrm{6.3\kern3mmEquation\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 5 − \frac{6}{x - 3} = 0\kern2mm\ } $$

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

             X-intercept, point A is (4,20 ; 0)
                                                                     [ Q 6.3 ]
    
$$ \hspace*{2 mm}\mathrm{6.4\kern3mmEquation\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 5 − \frac{6}{0 - 3}\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 7\kern2mm\ } $$
             Y-intercept, point B is (0 ; 7)
                                                                     [ Q 6.4 ]
    
$$ \hspace*{2 mm}\mathrm{6.5\kern3mmEquation\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{At\ D(5;d)\ :\ d = 5 − \frac{6}{5 - 3}\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{= 2\kern2mm\ } $$
                                                                     [ Q 6.5 ]
    
$$ \hspace*{2 mm}\mathrm{6.6\kern3mmEquation\ is\ y = 5 − \frac{6}{x - 3}\kern2mm\ } $$

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

                                                                     [ Q 6.6 ]
    
     6.7  At R and S the equations are equal,
            so that
$$ \hspace*{27 mm}\mathrm{5 − \frac{6}{x − 3} = − 2x + 15\kern2mm\ } $$

$$ \hspace*{6 mm}\mathrm{X (x − 3)\ \ :\ 5(x − 3) − 6 = (− 2x + 15)(x − 3)\kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{5x − 15 − 6 = −2x^2 + 21x − 45\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{x^2 − 8x + 12 = 0\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{(x − 2)(x − 6) = 0\kern2mm\ } $$
                                        x = 2  or  x = 6
                                      y = 11  or  y = 3
            The points of intersection are
            (2 ; 11) and (6 ; 3)
                                                                      [ Q 6.7 ]

  
Answers  7
    
     7.1  P is the point (− 1 ; − 5)                   [ Q 7.1 ]
    
$$ \hspace*{2 mm}\mathrm{7.2\kern3mmEquation\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

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

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

             X-intercept, point A is (− 2,67 ; 0)
                                                                     [ Q 7.2 ]
    
$$ \hspace*{2 mm}\mathrm{7.3\kern3mmEquation\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \frac{−5}{0 + 1} − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= − 8\kern2mm\ } $$
             Y-intercept, point B is (0 ; − 8)
                                                                     [ Q 7.3 ]
    
$$ \hspace*{2 mm}\mathrm{7.4\kern3mmEquation\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

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

$$ \hspace*{36 mm}\mathrm{= − 6,33\kern2mm\ } $$        [ Q 7.4 ]
    
$$ \hspace*{2 mm}\mathrm{7.5\kern3mmEquation\ is\ y = \frac{−5}{x + 1} − 3\kern2mm\ } $$

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

                                                                     [ Q 7.5 ]
    
     7.6  Positive axis of symmetry : y = (x + p) + q
                                                                 = (x + 1) − 5
                                                                 = x − 4
             Negative axis of symmetry : y = -(x + p) + q
                                                                    = −(x + 1) − 5
                                                                    = − x − 6
                                                                      [ Q 7.6 ]
  
Answers  8
    
$$ \hspace*{2 mm}\mathrm{8.1\kern3mmEquation\ is\ y = \frac{4}{x − 1} + 2\kern2mm\ } $$

            Horizontal asymptote : y = 2
            Vertical asymptote : x = 1
            The pieces that fit are p and r
$$ \hspace*{10 mm}\mathrm{Equation\ is\ y = \frac{4}{x + 2} − 2\kern2mm\ } $$

            Horizontal asymptote : y = − 2
            Vertical asymptote : x = − 2
            The pieces that fit are q and s
                                                                      [ Q 8.1 ]
    
$$ \hspace*{2 mm}\mathrm{8.2\kern3mmEquation\ is\ y = \frac{4}{x − 1} + 2 = \frac{4}{x − 1 + 3} + ( − 4)\kern2mm\ } $$

$$ \hspace*{34 mm}\mathrm{Graph A\ \ = \frac{4}{x + 2} − 2\kern2mm\ } $$
$$ \hspace*{53 mm}\mathrm{Graph B\kern2mm\ } $$
            Graph A is translated 3 units to
            the left and 4 units downwards
            to form graph B.
                                                                      [ Q 8.2 ]