Question

both types: find vertical and horizontal asymptotes

1. \\( \frac { 2 x + 1 } { x - 5 } \\) v:___ h:___
2. \\( \frac { 3 x + 4 } { x } \\) v:___ h:___
3. \\( \frac { x - 2 } { x + 3 } \\) v:___ h:___

vertical asymptotes: find where the denominator = 0

1. \\( \frac { 2 } { x - 3 } \\)_____
2. \\( \frac { x + 1 } { x + 4 } \\)_____
3. \\( \frac { 5 x } { 2 x - 8 } \\)_____
4. \\( \frac { x ^ { 2 } } { x ^ { 2 } - 9 } \\)_____

horizontal asymptotes: rewrite \\( \frac { a x + b } { x } = a + \frac { b } { x } \\)

1. \\( \frac { 3 x + 7 } { x } \\)_____
2. \\( \frac { 6 x - 2 } { x } \\)_____
3. \\( \frac { x + 8 } { x } \\)_____
4. \\( \frac { 4 x - 9 } { x } \\)_____

both types: find vertical and horizontal asymptotes

1. \\( \frac { 2 x + 1 } { x - 5 } \\) v:___ h:___
2. \\( \frac { 3 x + 4 } { x } \\) v:___ h:___
3. \\( \frac { x - 2 } { x + 3 } \\) v:___ h:___

context problem: average cost \\( c ( x ) = \frac { 50 + 3 x } { x } \\)

1. vertical asymptote:___ (what does this mean?)___
2. rewrite: \\( c ( x ) = \\)___ horizontal asymptote:___
3. as production increases, cost approaches $_____

Explanation:

Step1: Find vertical asymptotes

For a rational function \(y = \frac{f(x)}{g(x)}\), vertical asymptotes occur at \(x\) - values where \(g(x)=0\) (assuming \(f(x)
eq0\) at those \(x\) - values).

• For \(\frac{2x + 1}{x-5}\), set \(x - 5=0\), so \(x = 5\).
• For \(\frac{3x + 4}{x}\), set \(x=0\).
• For \(\frac{x - 2}{x + 3}\), set \(x+3=0\), so \(x=-3\).

Step2: Find horizontal asymptotes

For a rational function \(y=\frac{ax + b}{cx + d}\), if the degree of the numerator \(n\) and the degree of the denominator \(m\) are equal (\(n=m\)), the horizontal asymptote is \(y=\frac{a}{c}\).

• For \(\frac{2x + 1}{x-5}\), \(a = 2\), \(c = 1\), so \(y = 2\).
• For \(\frac{3x + 4}{x}\), rewrite as \(\frac{3x+4}{x}=3+\frac{4}{x}\), as \(x\to\pm\infty\), \(\frac{4}{x}\to0\), so \(y = 3\).
• For \(\frac{x - 2}{x + 3}\), \(a = 1\), \(c = 1\), so \(y = 1\).

Step3: Context problem

• Vertical asymptote: For \(C(x)=\frac{50 + 3x}{x}\), set \(x = 0\). The vertical asymptote \(x = 0\) means that the average - cost function is not defined at \(x = 0\) (you can't produce \(0\) units).
• Rewrite: \(C(x)=\frac{50+3x}{x}=3+\frac{50}{x}\). For horizontal asymptote, as \(x\to\pm\infty\), \(\frac{50}{x}\to0\), so \(y = 3\).
• Cost approach: As \(x\) (production) increases, \(C(x)\) approaches \(3\), so the cost approaches \(\$3\).

Answer:

1. \(\frac{2x + 1}{x-5}\): \(V:x = 5\), \(H:y = 2\)
2. \(\frac{3x + 4}{x}\): \(V:x = 0\), \(H:y = 3\)
3. \(\frac{x - 2}{x + 3}\): \(V:x=-3\), \(H:y = 1\)

Context problem:

1. \(V:x = 0\) (function not defined at \(x = 0\))
2. \(C(x)=3+\frac{50}{x}\), \(H:y = 3\)
3. \(3\)