-
\(f(x) = \dfrac{x}{3x - 6}\)
Domain: \((-\infty, 2) \cup (2, \infty)\)
Vertical asymptote: \(x = 2\)
As \(x \rightarrow 2^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 2^{+}, f(x) \rightarrow \infty\)
No holes in the graph
Horizontal asymptote: \(y = \frac{1}{3}\)
As \(x \rightarrow -\infty, f(x) \rightarrow \frac{1}{3}^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow \frac{1}{3}^{+}\)
-
\(f(x) = \dfrac{3 + 7x}{5 - 2x}\)
Domain: \((-\infty, \frac{5}{2}) \cup (\frac{5}{2}, \infty)\)
Vertical asymptote: \(x = \frac{5}{2}\)
As \(x \rightarrow \frac{5}{2}^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow \frac{5}{2}^{+}, f(x) \rightarrow -\infty\)
No holes in the graph
Horizontal asymptote: \(y = -\frac{7}{2}\)
As \(x \rightarrow -\infty, f(x) \rightarrow -\frac{7}{2}^{+}\)
As \(x \rightarrow \infty, f(x) \rightarrow -\frac{7}{2}^{-}\)
-
\(f(x) = \dfrac{x}{x^{2} + x - 12} = \dfrac{x}{(x + 4)(x - 3)}\)
Domain: \((-\infty, -4) \cup (-4, 3) \cup (3, \infty)\)
Vertical asymptotes: \(x = -4, x = 3\)
As \(x \rightarrow -4^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow -4^{+}, f(x) \rightarrow \infty\)
As \(x \rightarrow 3^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 3^{+}, f(x) \rightarrow \infty\)
No holes in the graph
Horizontal asymptote: \(y = 0\)
As \(x \rightarrow -\infty, f(x) \rightarrow 0^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow 0^{+}\)
-
\(f(x) = \dfrac{x}{x^{2} + 1}\)
Domain: \((-\infty, \infty)\)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: \(y = 0\)
As \(x \rightarrow -\infty, f(x) \rightarrow 0^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow 0^{+}\)
-
\(f(x) = \dfrac{x + 7}{(x + 3)^{2}}\)
Domain: \((-\infty, -3) \cup (-3, \infty)\)
Vertical asymptote: \(x = -3\)
As \(x \rightarrow -3^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow -3^{+}, f(x) \rightarrow \infty\)
No holes in the graph
Horizontal asymptote: \(y = 0\)
19As \(x \rightarrow -\infty, f(x) \rightarrow 0^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow 0^{+}\)
-
\(f(x) = \dfrac{x^{3} + 1}{x^{2} - 1} = \dfrac{x^{2} - x+ 1}{x-1}\)
Domain: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\)
Vertical asymptote: \(x = 1\)
As \(x \rightarrow 1^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 1^{+}, f(x) \rightarrow \infty\)
Hole at \((-1, -\frac{3}{2})\)
Slant asymptote: \(y=x\)
As \(x \rightarrow -\infty\), the graph is below \(y=x\)
As \(x \rightarrow \infty\), the graph is above \(y=x\)
-
\(f(x) = \dfrac{4x}{x^{2} + 4}\)
Domain: \((-\infty, \infty)\)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: \(y = 0\)
As \(x \rightarrow -\infty, f(x) \rightarrow 0^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow 0^{+}\)
-
\(f(x) = \dfrac{4x}{x^{2} -4} = \dfrac{4x}{(x + 2)(x - 2)}\)
Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)
Vertical asymptotes: \(x = -2, x = 2\)
As \(x \rightarrow -2^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow -2^{+}, f(x) \rightarrow \infty\)
As \(x \rightarrow 2^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 2^{+}, f(x) \rightarrow \infty\)
No holes in the graph
Horizontal asymptote: \(y = 0\)
As \(x \rightarrow -\infty, f(x) \rightarrow 0^{-}\)
As \(x \rightarrow \infty, f(x) \rightarrow 0^{+}\)
-
\(f(x) = \dfrac{x^2-x-12}{x^{2} +x - 6} = \dfrac{x-4}{x - 2}\)
Domain: \((-\infty, -3) \cup (-3, 2) \cup (2, \infty)\)
Vertical asymptote: \(x = 2\)
As \(x \rightarrow 2^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow 2^{+}, f(x) \rightarrow -\infty\)
Hole at \(\left(-3, \frac{7}{5} \right)\)
Horizontal asymptote: \(y = 1\)
As \(x \rightarrow -\infty, f(x) \rightarrow 1^{+}\)
As \(x \rightarrow \infty, f(x) \rightarrow 1^{-}\)
-
\(f(x) = \dfrac{3x^2-5x-2}{x^{2} -9} = \dfrac{(3x+1)(x-2)}{(x + 3)(x - 3)}\)
Domain: \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\)
Vertical asymptotes: \(x = -3, x = 3\)
As \(x \rightarrow -3^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow -3^{+}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 3^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 3^{+}, f(x) \rightarrow \infty\)
No holes in the graph
Horizontal asymptote: \(y = 3\)
As \(x \rightarrow -\infty, f(x) \rightarrow 3^{+}\)
As \(x \rightarrow \infty, f(x) \rightarrow 3^{-}\)
-
\(f(x) = \dfrac{x^3+2x^2+x}{x^{2} -x-2} = \dfrac{x(x+1)}{x - 2}\)
Domain: \((-\infty, -1) \cup (-1, 2) \cup (2, \infty)\)
Vertical asymptote: \(x = 2\)
As \(x \rightarrow 2^{-}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 2^{+}, f(x) \rightarrow \infty\)
Hole at \((-1,0)\)
Slant asymptote: \(y=x+3\)
As \(x \rightarrow -\infty\), the graph is below \(y=x+3\)
As \(x \rightarrow \infty\), the graph is above \(y=x+3\)
-
\(f(x) = \dfrac{x^3-3x+1}{x^2+1}\)
Domain: \((-\infty, \infty)\)
No vertical asymptotes
No holes in the graph
Slant asymptote: \(y=x\)
As \(x \rightarrow -\infty\), the graph is above \(y=x\)
As \(x \rightarrow \infty\), the graph is below \(y=x\)
-
\(f(x) = \dfrac{2x^{2} + 5x - 3}{3x + 2}\)
Domain: \(\left(-\infty, -\frac{2}{3}\right) \cup \left(-\frac{2}{3}, \infty\right)\)
Vertical asymptote: \(x = -\frac{2}{3}\)
As \(x \rightarrow -\frac{2}{3}^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow -\frac{2}{3}^{+}, f(x) \rightarrow -\infty\)
No holes in the graph
Slant asymptote: \(y = \frac{2}{3}x + \frac{11}{9}\)
As \(x \rightarrow -\infty\), the graph is above \(y = \frac{2}{3}x + \frac{11}{9}\)
As \(x \rightarrow \infty\), the graph is below \(y = \frac{2}{3}x + \frac{11}{9}\)
-
\(f(x) = \dfrac{-x^{3} + 4x}{x^{2} - 9} = \dfrac{-x^{3} + 4x}{(x-3)(x+3)}\)
Domain: \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\)
Vertical asymptotes: \(x = -3\), \(x=3\)
As \(x \rightarrow -3^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow -3^{+}, f(x) \rightarrow -\infty\)
As \(x \rightarrow 3^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow 3^{+}, f(x) \rightarrow -\infty\)
No holes in the graph
Slant asymptote: \(y=-x\)
As \(x \rightarrow -\infty\), the graph is above \(y=-x\)
As \(x \rightarrow \infty\), the graph is below \(y=-x\)
-
\(f(x) = \dfrac{-5x^{4} - 3x^{3} + x^{2} - 10}{x^{3} - 3x^{2} + 3x - 1} \\ \phantom{f(x)} = \dfrac{-5x^{4} - 3x^{3} + x^{2} - 10}{(x-1)^3}\)
Domain: \((-\infty, 1) \cup (1, \infty)\)
Vertical asymptotes: \(x = 1\)
As \(x \rightarrow 1^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow 1^{+}, f(x) \rightarrow -\infty\)
No holes in the graph
Slant asymptote: \(y=-5x-18\)
As \(x \rightarrow -\infty\), the graph is above \(y=-5x-18\)
As \(x \rightarrow \infty\), the graph is below \(y=-5x-18\)
-
\(f(x) = \dfrac{x^3}{1-x}\)
Domain: \((-\infty, 1) \cup (1, \infty)\)
Vertical asymptote: \(x=1\)
As \(x \rightarrow 1^{-}, f(x) \rightarrow \infty\)
As \(x \rightarrow 1^{+}, f(x) \rightarrow -\infty\)
No holes in the graph
No horizontal or slant asymptote
As \(x \rightarrow -\infty\), \(f(x) \rightarrow -\infty\)
As \(x \rightarrow \infty\), \(f(x) \rightarrow -\infty\)
-
\(f(x) = \dfrac{18-2x^2}{x^2-9} = -2\)
Domain: \((-\infty, -3) \cup (-3,3) \cup (3, \infty)\)
No vertical asymptotes
Holes in the graph at \((-3,-2)\) and \((3,-2)\)
Horizontal asymptote \(y = -2\)
As \(x \rightarrow \pm \infty\), \(f(x) = -2\)
-
\(f(x) = \dfrac{x^3-4x^2-4x-5}{x^2+x+1} = x-5\)
Domain: \((-\infty, \infty)\)
No vertical asymptotes
No holes in the graph
Slant asymptote: \(y = x-5\)
\(f(x) = x-5\) everywhere.
-
- \(C(25) = 590\) means it costs $590 to remove 25% of the fish and and \(C(95)= 33630\) means it would cost $33630 to remove 95% of the fish from the pond.
- The vertical asymptote at \(x = 100\) means that as we try to remove 100% of the fish from the pond, the cost increases without bound; i.e., it’s impossible to remove all of the fish.
- For $40000 you could remove about 95.76% of the fish.
- The horizontal asymptote of the graph of \(P(t) = \frac{150t}{t + 15}\) is \(y = 150\) and it means that the model predicts the population of Sasquatch in Portage County will never exceed 150.
-
- \(\overline{C}(x) = \frac{100x+2000}{x}\), \(x > 0\).
- \(\overline{C}(1) = 2100\) and \(\overline{C}(100) = 120\). When just \(1\) dOpi is produced, the cost per dOpi is \(\$2100\), but when \(100\) dOpis are produced, the cost per dOpi is \(\$120\).
- \(\overline{C}(x) = 200\) when \(x = 20\). So to get the cost per dOpi to \(\$200\), \(20\) dOpis need to be produced.
- As \(x \rightarrow 0^{+}\), \(\overline{C}(x) \rightarrow \infty\). This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded. In this situation, there is a fixed cost of \(\$2000\) (\(C(0) = 2000\)), we are trying to spread that \(\$2000\) over fewer and fewer dOpis.
- As \(x \rightarrow \infty\), \(\overline{C}(x) \rightarrow 100^{+}\). This means that as more and more dOpis are produced, the cost per dOpi approaches \(\$100\), but is always a little more than \(\$100\). Since \(\$100\) is the variable cost per dOpi (\(C(x) = \underline{100}x+2000\)), it means that no matter how many dOpis are produced, the average cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to the \(\$2000\) fixed cost, which factors into the average cost per dOpi no matter how many dOpis are produced.
-
- The maximum power is approximately \(1.603 \; mW\) which corresponds to \(3.9 \; k\Omega\).
- As \(x \rightarrow \infty, \; P(x) \rightarrow 0^{+}\) which means as the resistance increases without bound, the power diminishes to zero.