MATHEMATICS
Grade 11
MORE EXERCISES
Graphs of the exponential function : answers.
  
  
Answers  1
    
$$ \hspace*{2 mm}\mathrm{1.1\kern3mmy = 3^{x+1}  − 1\kern2mm\ } $$

           Horizontal asymptote : y = − 1
           Base > 1, function is increasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 3^{0+1}  − 1\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 3 − 1\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2\kern2mm\ } $$
             Y-intercept is (0 ; 2)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 3^{x+1}  − 1 = 0\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{3^{x+1} = 1 = 3^0\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{x + 1 = 0\kern2mm\ } $$
$$ \hspace*{47 mm}\mathrm{x = − 1\kern2mm\ } $$

             X-intercept is (−1 ; 0)
  

                                                                   [ Q 1.1 ]
    
$$ \hspace*{2 mm}\mathrm{1.2\kern3mmy = 3^{x − 1}   + 2\kern2mm\ } $$

           Horizontal asymptote : y = 2
           Base > 1, function is increasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 3^{0 − 1}   + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 0,33 + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2,33\kern2mm\ } $$
             Y-intercept is (0 ; 2,33)
             No X-intercept because q > 0
  

                                                                   [ Q 1.2 ]
    
$$ \hspace*{2 mm}\mathrm{1.3\kern3mmy = 2^{1 − x}   + 3\kern2mm\ } $$

           Horizontal asymptote : y = 3
           Base < 1, function is decreasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 2^{1 − 0}   + 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2 + 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 5\kern2mm\ } $$
             Y-intercept is (0 ; 5)
             No X-intercept because q > 0
  

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

           Horizontal asymptote : y = − 3
           Base < 1, function is decreasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 3^{2 − 0}  − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 9 − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 6\kern2mm\ } $$
             Y-intercept is (0 ; 6)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 3^{2 − x}  − 3 = 0\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{3^{2 − x} = 3^1\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{2 − x = 1\kern2mm\ } $$
$$ \hspace*{48 mm}\mathrm{x = 1\kern2mm\ } $$

             X-intercept is (1 ; 0)
  

                                                                   [ Q 1.4 ]
    
$$ \hspace*{2 mm}\mathrm{1.5\kern3mmy = 2.3^{x − 1}  + 2\kern2mm\ } $$

           Horizontal asymptote : y = 2
           Base > 1, function is increasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 2.3^{0 − 1}  + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 0,67 + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2,67\kern2mm\ } $$
             Y-intercept is (0 ; 2,67)
             No X-intercept because q > 0
  

                                                                   [ Q 1.5 ]
    
$$ \hspace*{2 mm}\mathrm{1.6\kern3mmy = 3.2^{1 − x}   − 3\kern2mm\ } $$

           Horizontal asymptote : y = − 3
           Base < 1, function is decreasing.
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 3.2^{1 − 0}   − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 6 − 3\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 3\kern2mm\ } $$
             Y-intercept is (0 ; 3)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 3.2^{1 − x}   − 3 = 0\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{3.2^{1 − x} = 3^1\kern2mm\ } $$
$$ \hspace*{47 mm}\mathrm{2^{1 − x} = 1 = 2^0\kern2mm\ } $$
$$ \hspace*{47 mm}\mathrm{1 − x = 0\kern2mm\ } $$
$$ \hspace*{53 mm}\mathrm{x = 1\kern2mm\ } $$

             X-intercept is (1 ; 0)
  

                                                                   [ Q 1.6 ]
  
Answers  2
    
   2.1  Horizontal asymptote : y = 1
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = 2^{x + p} + 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ B(0;1,5)\ :\ 2^{0 + p} + 1 = 1,5\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{2^p = 0,5\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{2^p = 2^− 1\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{p = − 1\kern2mm\ } $$
           p = − 1 and q = 1
$$ \hspace*{9 mm}\mathrm{y = 2^{x − 1} + 1\kern2mm\ } $$                                 [ Q 2.1 ]
    
   2.2  Horizontal asymptote : y = − 3
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = \Big(\frac{1}{3}\Big)^{x + p} + q\kern2mm\ } $$
$$ \hspace*{28mm}\mathrm{y = 3^{−(x + p)} − 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ A(1;0)\ :\ 3^{−(1 + p)} − 3 = 0\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{3^{−(1 + p)} = 3\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{−(1 + p) = 1\kern2mm\ } $$
$$ \hspace*{41 mm}\mathrm{p = − 2\kern2mm\ } $$
           p = − 1 and q = 1
$$ \hspace*{9 mm}\mathrm{y = 2^{x − 1} + 1\kern2mm\ } $$                                 [ Q 2.2 ]
    
   2.3  Horizontal asymptote : y = − 2, thus q = − 2
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = 2^{x + p} − 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ A(3;0)\ :\ 2^{3 + p} − 2 = 0\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{2^{3 + p} = 2\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 + p = 1\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{p = − 2\kern2mm\ } $$
           p = − 2 and q = − 2
$$ \hspace*{9 mm}\mathrm{y = 2^{x − 2} − 2\kern2mm\ } $$                                 [ Q 2.3 ]
    
   2.4  Horizontal asymptote : y = 2, thus q = 2
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = \Big(\frac{1}{2}\Big)^{x + p} + q\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{y = \Big(\frac{1}{2}\Big)^{x + p} + 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{y = 2^{−(x + p)} + 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ B(0;2,33)\ :\ 2^{0 + p} − 2 = 2,33\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{2^{3 + p} = 2\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{3 + p = 1\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{p = − 2\kern2mm\ } $$
           p = − 2 and q = − 2
$$ \hspace*{9 mm}\mathrm{y = 2^{x − 2} − 2\kern2mm\ } $$                                 [ Q 2.4 ]
    
   2.5  Horizontal asymptote : y = 1, thus q = 1
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = a.b^x + 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ B(0;3)\ :\ a.b^{0} + 1 = 3\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{a.1 + 1 = 3\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{a = 2\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = 2.b^x + 1\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ A(1;7)\ :\ 2.b^{1} + 1 = 7\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{b = 3\kern2mm\ } $$
           a = 2, b = 3 and q = 1
$$ \hspace*{9 mm}\mathrm{y = 2.3^x + 1\kern2mm\ } $$                                     [ Q 2.5 ]
    
   2.6  Horizontal asymptote : y = − 3, thus q = − 3
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = a.b^x − 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ A(0;0)\ :\ a.b^{0} − 3 = 0\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{a.1 − 3 = 0\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{a = 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{Equation\ :\ y = 3.b^x − 3\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{At\ B(−2;9)\ :\ 3.b^{−2} − 3 = 9\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{3.b^{−2} = 12\kern2mm\ } $$
$$ \hspace*{9 mm}\mathrm{b^{−2} = 4 = (2^{−1})^{−2}\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{b = 2^{−1}\kern2mm\ } $$
           a = 3, b = ½ and q = − 3
$$ \hspace*{9 mm}\mathrm{y = 3.\Big(\frac{1}{2}\Big)^x − 3\kern2mm\ } $$                               [ Q 2.6 ]

  
Answers  3
    
$$ \hspace*{2 mm}\mathrm{3.1\kern3mmy = 4^{x − 1}  − 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 4^{0 − 1}  − 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 4^{ − 1}  − 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= \frac{1}{4}  − 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= − 1,75\kern2mm\ } $$
             Y-intercept is (0 ; − 1,75)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 4^{x − 1}  − 2 = 0\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{4^{x − 1} = 2\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{2^{2(x − 1)} = 2^1\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{2(x − 1) = 1\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{2x − 2 = 1\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{x = 1,5\kern2mm\ } $$
             X-intercept is (1,5 ; 0)                     [ Q 3.1 ]
    
$$ \hspace*{2 mm}\mathrm{3.2\kern3mmy = 4^{x − 1}  − 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ P(2;p) \ :\ p = 4^{2 − 1}  − 2\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{p = 4^{1} − 2\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{= 2\kern2mm\ } $$                      [ Q 3.2 ]
    
$$ \hspace*{2 mm}\mathrm{3.3\kern3mmy = 4^{x − 1}  − 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ R(r ; −1) \ :\ −1= 4^{r − 1}  − 2\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{1 = 4^{r − 1}\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{4^0 = 4^{r − 1}\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{0 = r − 1\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{r = 1\kern2mm\ } $$                      [ Q 3.3 ]
    
  3.4   4x − 1   − 2 < 0, i.e. negative, below
          the X-axis. Thus x < X-intercept.
          Thus 4x − 1   − 2 < 0, if x < 1,5             [ Q 3.4 ]
    
   3.5  Domain : x = {x : x ∈ ℜ}                  [ Q 3.5 ]
    
   3.6  Range : y = {y : y > −2; y ∈ ℜ}         [ Q 3.6 ]
  
Answers  4
    
$$ \hspace*{2 mm}\mathrm{4.1\kern3mmy = 2^{−x + 1}   + 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 2^{0 + 1}   + 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2 + 2 \kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 4\kern2mm\ } $$
             No X-intercept because y = 2
                                                                       [ Q 4.1 ]
    
$$ \hspace*{2 mm}\mathrm{4.2\kern3mmy = 2^{−x + 1}   + 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ P(−1;p) \ :\ p = 2^{−(−1) + 1}   + 2\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{= 2^2 + 2\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{= 6\kern2mm\ } $$                      [ Q 4.2 ]
    
$$ \hspace*{2 mm}\mathrm{4.3\kern3mmy = 2^{−x + 1}   + 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ R(r ; 2,25) \ :\ 2^{−r + 1}   + 2 = 2,25\kern2mm\ } $$
$$ \hspace*{44 mm}\mathrm{2^{−r + 1} = 0,25\kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{2^{−r + 1} = 4^{− 1} = 2^{−2}\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{−r + 1 = −2\kern2mm\ } $$
$$ \hspace*{49 mm}\mathrm{r = 3\kern2mm\ } $$
                                                                        [ Q 4.3 ]
    
   4.4  Domain : x = {x : x ∈ ℜ}                  [ Q 4.4 ]
    
   4.5  Range : y = {y : y > 2; y ∈ ℜ}           [ Q 4.5 ]
    
   4.6  Function is positive for
          all values of x.                                  [ Q 4.6 ]
    
$$ \hspace*{2 mm}\mathrm{4.7\kern3mmh(x) = 2^{−x + 1}   + (2 − 4)\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{ = 2^{−x + 1}   − 2\kern2mm\ } $$                    [ Q 4.7 ]
  
Answers  5
    
$$ \hspace*{2 mm}\mathrm{5.1\kern3mmy = \Big(\frac{1}{2}\Big)^{x − 1} − 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = \Big(\frac{1}{2}\Big)^{0 − 1} − 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= \Big(\frac{1}{2}\Big)^{− 1} − 4 \kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= (2^{−1})^{− 1} − 4 \kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 2 − 4 \kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −2\kern2mm\ } $$
             Y-intercept is (0 ; − 2)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ \Big(\frac{1}{2}\Big)^{x − 1} − 4 = 0\kern2mm\ } $$
$$ \hspace*{40 mm}\mathrm{(2^{−1})^{x− 1} = 2^2 \kern2mm\ } $$
$$ \hspace*{45 mm}\mathrm{2^{−x + 1} = 2^2 \kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{−x + 1 = 2 \kern2mm\ } $$
$$ \hspace*{50 mm}\mathrm{x = − 1 \kern2mm\ } $$
             X-intercept is (−1 ; 0)                     [ Q 5.1 ]
    
$$ \hspace*{2 mm}\mathrm{5.2\kern3mmy = \Big(\frac{1}{2}\Big)^{x − 1} − 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ C(c;4) \ :\ 4 = \Big(\frac{1}{2}\Big)^{x − 1} − 4\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{8 = (2^{−1})^{x− 1}\kern2mm\ } $$
$$ \hspace*{27 mm}\mathrm{2^3 = 2^{−x + 1}\kern2mm\ } $$ $$ \hspace*{28 mm}\mathrm{3 = −x + 1 \kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{x = − 2 \kern2mm\ } $$                      [ Q 5.2 ]
    
   5.3  Domain : x = {x : x ∈ ℜ}                    [ Q 5.3 ]
    
   5.4  Range : y = {y : y > − 4; y ∈ ℜ}           [ Q 5.4 ]
    
$$ \hspace*{2 mm}\mathrm{5.5\kern3mm\Big(\frac{1}{2}\Big)^{x − 1} − 4 > 0\ if\ x < − 1\kern2mm\ } $$            [ Q 5.5 ]
    
$$ \hspace*{2 mm}\mathrm{5.6\kern3mmh(x) = \Big(\frac{1}{2}\Big)^{x − 1} − (4 + 2)\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{ = \Big(\frac{1}{2}\Big)^{x − 1}  − 2\kern2mm\ } $$                    [ Q 5.6 ]
  
Answers  6
    
  6.1   Horizontal asymptote : y = − 1 and q = − 1
$$ \hspace*{5 mm}\mathrm{\kern3mmy = \Big(\frac{1}{3}\Big)^{x + p} + q\kern2mm\ } $$

$$ \hspace*{5 mm}\mathrm{\kern3mmAt\ B(0;8)\ :\ \ 8 = \Big(\frac{1}{3}\Big)^{0 + p} − 1\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{9 = (3^{−1})^p\kern2mm\ } $$
$$ \hspace*{26 mm}\mathrm{3^2 = 3^{−p}\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{2 = −p\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{p = −2\kern2mm\ } $$
             p = − 2 en q = − 1                             [ Q 6.1 ]
    
$$ \hspace*{2 mm}\mathrm{6.2\kern3mmy = \Big(\frac{1}{3}\Big)^{x − 2} − 1\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ P(p;2) \ :\ 2 = \Big(\frac{1}{3}\Big)^{p − 2} − 1\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3 = (3^{−1})^{p− 2}\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{3 = 3^{−p + 2}\kern2mm\ } $$ $$ \hspace*{29 mm}\mathrm{1 = −p + 2 \kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{p = 1\kern2mm\ } $$                      [ Q 6.2 ]
    
   6.3  Domain : x = {x : x ∈ ℜ}                    [ Q 6.3 ]
    
   6.4  Range : y = {y : y > − 1; y ∈ ℜ}           [ Q 6.4 ]
    
$$ \hspace*{2 mm}\mathrm{6.5\kern3mmh(x) = \Big(\frac{1}{3}\Big)^{x − 2} − 1 + 3\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{ = \Big(\frac{1}{3}\Big)^{x − 2} + 2\kern2mm\ } $$                    [ Q 6.5 ]
    
   6.6  The graph is translated 3 units to
           the top and thus all points are
           translated 3 units to the top.
           The Y-intercept of h(x) will be 8 + 3 = 11
           The Y-intercept of h(x) will be (0 ; 11)
                                                                       [ Q 6.6 ]
  
Answers  7
    
  7.1   Horizontal asymptote : y = − 2       [ Q 7.1 ]
    
$$ \hspace*{2 mm}\mathrm{7.2\kern3mmy = 2^x − 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Y-intercept\ \ :\ y = 2^0 − 2\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −1 \kern2mm\ } $$
             Y-intercept is (0 ; − 1)
$$ \hspace*{9 mm}\mathrm{X-intercept\ \ :\ 2^x − 2 = 0\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{2^x = 2 \kern2mm\ } $$
$$ \hspace*{43 mm}\mathrm{x = 1 \kern2mm\ } $$
             X-intercept is (1 ; 0)                       [ Q 7.2 ]
    
$$ \hspace*{2 mm}\mathrm{7.3\kern3mmAt\ P(p;30)\ :\ \ 30 = 2^p − 2\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{32 = 2^p\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{2^5 = 2^p\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{5 = p\kern2mm\ } $$                     [ Q 7.3 ]
    
  7.4   h(x) ≤ 30 if x ≤ 5                               [ Q 7.4 ]
  
Answers  8
    
  8.1   Horizontal asymptote : y = 4       [ Q 8.1 ]
    
$$ \hspace*{2 mm}\mathrm{8.2\kern3mmy = p.2^{−x}  + q\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{q = 4\ \kern3mm\ . . . horizontal\ \ asymptote\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{\therefore\ y = p.2^{−x}  + 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ D(−4 ;−12)\ : \ \ −12 = p.2^{−(−4)}  + 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{−16 = p.2^4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{−16 = p.16\kern2mm\ } $$
$$ \hspace*{42 mm}\mathrm{p = −1 \kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{\therefore\ f(x) = −2^{−x}  + 4\kern2mm\ } $$
                                                                          [ Q 8.2 ]
    
$$ \hspace*{2 mm}\mathrm{8.3\kern3mm f(x) = −2^{−x}  + 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ E(2;e)\ : \ \ e = −2^{−2}  + 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= −\frac{1}{4} + 4\kern2mm\ } $$
$$ \hspace*{38 mm}\mathrm{= 3,75\kern2mm\ } $$               [ Q 8.3 ]
      $$ \hspace*{2 mm}\mathrm{8.4\kern3mm f(x) < 3,75\ \ if\ \ x < 2\kern2mm\ } $$                         [ Q 8.4 ]
     $$ \hspace*{2 mm}\mathrm{8.5\kern3mm f(x) > 0\ i.e.\ positive\ if\ it\ is\ above\ the\ X-axis.\kern2mm\ } $$
            and therefore x > X-intercept.
            Thus calculate the X-intercept.
$$ \hspace*{24 mm}\mathrm{0 = −2^{−x}  + 4\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{2^{−x}  = 2^2\kern2mm\ } $$
$$ \hspace*{21 mm}\mathrm{−x  = 2\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{x  = −2\kern2mm\ } $$
$$ \hspace*{20 mm}\mathrm{f(x) > 0\ \ if\ \ x > −2\kern2mm\ } $$                [ Q 8.5 ]
  
Answers  9
    
$$ \hspace*{2 mm}\mathrm{9.1\kern3mmg(x) = 2^{x + p}  + q\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Y-intercept\ :\ −1 = 2^{0 + p}  + q\kern2mm\ } $$
$$ \hspace*{34 mm}\mathrm{−1 = 2^p  + q\ \kern3mm\ (1)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{At\ A(2 ;11)\ : \ \ 11 = 2^{2 + p}  + q\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{11 = 2^2 \times 2^p  + q\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{11 = 4 \times 2^p  + q\ \kern3mm\ (2)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{(2) − (1)\ : \ \kern5mm\ 12 = 3 \times 2^p\kern2mm\ } $$
$$ \hspace*{35 mm}\mathrm{4 = 2^p\kern2mm\ } $$
$$ \hspace*{35 mm}\mathrm{2 = p\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{Into (1)\ : \ \kern6mm\ −1 = 2^2  + q\kern2mm\ } $$
$$ \hspace*{32 mm}\mathrm{−5 = q\kern2mm\ } $$
$$ \hspace*{30 mm}\mathrm{g(x) = 2^{x + 2}  − 5\kern2mm\ } $$
                                                                          [ Q 9.1 ]
    
$$ \hspace*{2 mm}\mathrm{9.2\ \ Horizontal\ asymptote\ \ :\ y = −5\kern2mm\ } $$
                                                                          [ Q 9.2 ]
    
$$ \hspace*{2 mm}\mathrm{9.3\ \ The\ graph\ is\ translated\ 4\ units\ downwards\kern2mm\ } $$
           and thus y changes to (y + 4).
$$ \hspace*{12 mm}\mathrm{\therefore g(x) = y = 2^{x + 2}  − 5\kern2mm\ } $$
$$ \hspace*{19 mm}\mathrm{y + 4 = 2^{x + 2}  − 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 2^{x + 2}  − 5− 4\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 2^{x + 2}  − 9\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 2^{x + 2}  − 9\kern2mm\ } $$                                 [ Q 9.3 ]
    
$$ \hspace*{2 mm}\mathrm{9.4\ \ The\ graph\ is\ translated\ 3\ units\ to\ the\ right\kern2mm\ } $$
           and thus x changes to (x − 3)
$$ \hspace*{12 mm}\mathrm{\therefore g(x) = y = 2^{x + 2}  − 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 2^{(x − 3) + 2}  − 5\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 2^{x − 1}  − 5\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{k(x) = 2^{x + 2}  − 5\kern2mm\ } $$                                 [ Q 9.4 ]
  
Answers  10
    
$$ \hspace*{2 mm}\mathrm{10.1\ \ The\ graph\ is\ translated\ 3\ units\ downwards\kern2mm\ } $$
             and thus y changes to (y + 3).
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = 2^x\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{y + 3 = 2^x\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 2^x  − 3\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 2^x  − 3\kern2mm\ } $$                                 [ Q 10.1 ]
    
$$ \hspace*{2 mm}\mathrm{10.2\ \ The\ graph\ is\ translated\ 2\ units\ to\ the\ left\kern2mm\ } $$
             and thus x changes to (x + 2)
$$ \hspace*{12 mm}\mathrm{\therefore g(x) = y = 2^x\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 2^{x + 2}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{k(x) = 2^{x + 2}\kern2mm\ } $$                                       [ Q 10.2 ]
  
Answers  11
    
$$ \hspace*{2 mm}\mathrm{11.1\ \ The\ graph\ is\ translated\ 2\ units\ upward\kern2mm\ } $$
             and thus y changes to (y − 2).
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = 3^{−x}\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{y − 2 = 3^{−x}\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 3^{−x}  + 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{g(x) = 3^{−x}  + 2\kern2mm\ } $$                                 [ Q 11.1 ]
    
$$ \hspace*{2 mm}\mathrm{11.2\ \ The\ graph\ is\ translated\ 4\ units\ to\ the\ right\kern2mm\ } $$
             and thus x changes to (x − 4)
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = 3^{−x}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 3^{−(x − 4)}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 3^{−x + 4}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 3^{−x + 4}\kern2mm\ } $$                                 [ Q 11.2 ]
$$ \hspace*{25 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 3^{−(x − 4)}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 3^{−x + 4}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 3^{−x} \times 3^4\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 81 \times 3^{−x}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 81 \times 3^{−x}\kern2mm\ } $$                                 [ Q 11.2 ]
  
Answers  12
    
$$ \hspace*{2 mm}\mathrm{12.1\ \ The\ graph\ is\ translated\ 4\ units\ downwards\kern2mm\ } $$
             and thus y changes to (y + 4).
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = 5^{x + 3}\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{y + 4 = 5^{x + 3}\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 5^{x + 3}  − 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{g(x) = 5^{x + 3}  − 4\kern2mm\ } $$                                 [ Q 12.1 ]
$$ \hspace*{25 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 5^{x + 3}  − 4\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 5^x \times 5^3  − 4\kern2mm\ } $$
$$ \hspace*{28 mm}\mathrm{y = 5^x \times 125  − 4\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{g(x) = 125 \times 5^x  − 4\kern2mm\ } $$                        [ Q 12.1 ]
    
$$ \hspace*{2 mm}\mathrm{12.2\ \ The\ graph\ is\ translated\ 4\ units\ to\ the\ left\kern2mm\ } $$
             and thus x changes to (x + 4)
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = 5^{x + 3}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 5^{(x + 4) + 3}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 5^{x + 7}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 5^{x + 7}\kern2mm\ } $$                                 [ Q 12.2 ]
$$ \hspace*{25 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 5^{x + 7}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 5^x \times 5^7\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = 78 125 \times 5^x\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = 78 125 \times 5^x\kern2mm\ } $$                           [ Q 12.2 ]
  
Answers  13
    
$$ \hspace*{2 mm}\mathrm{13.1\ \ The\ graph\ is\ translated\ 2\ units\ upward\kern2mm\ } $$
             and thus y changes to (y − 2).
$$ \hspace*{15 mm}\mathrm{\therefore f(x) = − 3^{−x + 2}\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{y − 2 = − 3^{−x + 2}\kern2mm\ } $$
$$ \hspace*{22 mm}\mathrm{y = − 3^{−x + 2} + 2\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{g(x) = − 3^{−x + 2}  + 2\kern2mm\ } $$                      [ Q 13.1 ]
    
$$ \hspace*{2 mm}\mathrm{13.2\ \ The\ graph\ is\ translated\ 3\ units\ to\ the\ right\kern2mm\ } $$
             and thus x changes to (x − 3)
$$ \hspace*{12 mm}\mathrm{\therefore f(x) = y = − 3^{−x + 2}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = − 3^{−(x−3) + 2}\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{y = − 3^{−x + 5}\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = − 3^{−x + 5}\kern2mm\ } $$                                 [ Q 13.2 ]