Answers


    1. \( f(x) \rightarrow 3 \) as \( x\rightarrow \pm\infty \) \( \left(\displaystyle \lim_{x \to \pm\infty} \frac{3x}{x-4}= 3 \right)\), \( \text{HA} \): \( y=3 \)
    2. \( g(x) \rightarrow 0 \) as \( x\rightarrow \pm\infty \) \( \left(\displaystyle \lim_{x \to \pm\infty} \frac{2x}{x^2+4}= 0 \right)\), \( \text{HA} \): \( y=0 \)
    3. \( h(x) \rightarrow \frac{1}{2} \) as \( x\rightarrow \pm\infty \) \(\left( \displaystyle \lim_{x \to \pm\infty} \frac{x^2}{2x^2-1}= \frac{1}{2} \right)\), \( \text{HA} \): \( y=\frac{1}{2} \)
    4. \( f(x) \rightarrow \infty \) as \( x\rightarrow \infty \) \( \left( \displaystyle \lim_{x \to \infty} \frac{x^2}{2x+1}= \infty \right) \), \( f(x) \rightarrow -\infty \) as \( x\rightarrow -\infty \) \( \left( \displaystyle \lim_{x \to -\infty} \frac{x^2}{2x+1}= -\infty \right) \), \( \text{HA} \): none
    1. \( \text{HA} \): \( y=2 \), \( f(x) \rightarrow 2 \) from above as \( x\rightarrow \infty \), \( f(x) \rightarrow 2 \) from below as \( x\rightarrow -\infty \)
    2. \( \text{HA} \): \( y=-1 \), \( g(x) \rightarrow -1 \) from above as \( x\rightarrow \pm\infty \)
    3. \( \text{HA} \): \( y=0 \), \( h(x) \rightarrow 0 \) from above as \( x\rightarrow \infty \), \( h(x) \rightarrow 0 \) from below as \( x\rightarrow -\infty \)
    1. The degree of the numerator is \( 1 \) greater than the degree of the denominator.
    2. Divide the denominator into the numerator and express the function in the form \( y=q(x)+\dfrac{r(x)}{d(x)} \). As \( x\rightarrow \pm\infty, \dfrac{r(x)}{d(x)} \rightarrow 0 \) since the degree of \( r(x) \) is less than the degree of \( d(x) \). Thus, \( y \rightarrow q(x) \) and \( y=q(x) \) is the oblique asymptote.
    3. (i) \( y= x -3 \), (iv) \( y=2x+5 \), (v) \( y=x \)
    1. \( \text{VA} \): \( x=-1, x=3 \); \( \text{HA} \): \( y=0 \)
    2. \( \text{VA} \): \( x=-1 \); \( \text{HA} \): \( y=\frac{1}{2} \)
    3. \( \text{VA} \): \( x=1 \); \( \text{OA} \): \( y=x+1 \)
    4. \( \text{VA} \): \( x=0, x=3 \); \( \text{HA} \): \( y=0 \)
    5. \( \text{VA} \): \( x=-3 \); Hole at \( (3, -\frac{1}{6}) \); \( \text{HA} \): \( y=1 \)
    6. \( \text{OA} \): \( y=-2x \)
  5. \( a=6, b=-2, c=8 \)
  6. \( a=72, b=36 \)
  7. \( n=1, a=2, b=7, c=3 \)
    1. \( $ 11 \ 555 \)
    2. \( \text{HA} \): \( S=130 \). Since the function approaches the horizontal asymptote from below, the company can expect to make no more than \( $13 \ 000 \) in weekly sales, no matter how much they spend on advertising.
    1. \( f(x)=2x+\frac{42}{x}, x \gt 0 \)
    2. \( \text{VA} \): \( x=0 \), \( \text{OA} \): \( y=2x \),
      \( f(x) \) \( x \)
      \( 1 \) \( 44 \)
      \( 2 \) \( 25 \)
      \( 3 \) \( 20 \)
      \( 4 \) \( 18.5 \)
      \( 5 \) \( 18.4 \)
      \( 6 \) \( 19 \)
      \( 7 \) \( 20 \)
      \( 8 \) \( 21.25 \)
      Oblique asymptote y=2x, vertical asymptote x=0, function positive for x>0 and minimum at (4.5,18.3)
    3. From the table of values, the minimum value occurs somewhere between \(4\) and \(6\), most likely between \(4\) and \(5\). \( f(4.5) \approx 18.3333 \) and checking each side of \(4.5\) we have,
      \( f(x) \) \( x \)
      \( 4.4 \) \( 18.3454 \)
      \( 4.5 \) \( 18.3333 \)
      \( 4.6 \) \( 18.3304 \)
      \( 4.7 \) \( 18.3362 \)
      The minimum amount of fencing is required when \( x\approx 4.6\) \(\left(y\approx \frac{21}{4.6}\approx 4.6 \right)\). Therefore the dimensions of the garden are \( 4.6~\text{m} \) by \( 4.6~\text{m} \).
    1. \( f(x)=x^2-1-\dfrac{3}{x-3} \)
    2. The graph of \( f(x)=\dfrac{x^3-3x^2-x}{x-3} \) will approach the parabola \( q(x)=x^2-1 \) more and more closely as \( x\rightarrow \pm\infty \), similar to the way the graph of a function would approach its horizontal or oblique asymptote.
      Graph of f(x) and q(x)=x^2-1; the curve f(x) resembles q(x) more and more as x goes to infinity
    3. As \( x\rightarrow \infty \), the graph of the function approaches \( y=\sqrt{x} \) more and more closely.
      Graph of (x + 1)/(sqrt(x)) and sqrt(x); (x + 1)/sqrt(x) resembles sqrt(x) more and more as x goes to infinity
  11. The sum is defined by \( y=x+\dfrac{1}{x}, x \gt 0 \), \( \text{VA} \): \( x=0 \), \( \text{OA} \): \( y=x \).
    \( x \) \( y \)
    \( 0.5 \) \( 2.5 \)
    \( 1 \) \( 2 \)
    \( 1.5 \) \( 2.17 \)
    \( 2 \) \( 2.5 \)
    \( 3 \) \( 3.33 \)
    From this table it appears that a minimum sum occurs between \(x=0.5\) and \(x=1.5\). Checking values of \(y\) on either side of \(x=1\):
    \( x \) \( y \)
    \( 0.5 \) \( 2.5 \)
    \( 0.9 \) \( 2.0111 \)
    \( 0.99 \) \( 2.0001 \)
    \( 1 \) \( 2 \)
    \( 1.01 \) \( 2.0001 \)
    \( 1.1 \) \( 2.0091 \)
    \( 1.5 \) \( 2.17 \)

    Oblique asymptote y=x, vertical asymptote x=0, function positive for x>0 and minimum at (1,2)

    When the number is \( 1 \), the sum is the smallest.