Question

1. sketch the graph of an example of a function f that satisfies all the following conditions. f(0) = 0, \lim_{x\to 1^-} f(x) = \infty, \lim_{x\to 1^+} f(x) = -\infty, \lim_{x\to -\infty} f(x) = -2, \lim_{x\to \infty} f(x) = -2

Explanation:

Step1: Analyze Vertical Asymptote

The left - hand limit as \(x\to1^{-}\) is \(\infty\) and the right - hand limit as \(x\to1^{+}\) is \(-\infty\), so there is a vertical asymptote at \(x = 1\).

Step2: Analyze Horizontal Asymptote

The limit as \(x\to-\infty\) and \(x\to\infty\) is \(- 2\), so there is a horizontal asymptote at \(y=-2\).

Step3: Analyze the Point \((0,0)\)

We know that \(f(0) = 0\), so the graph passes through the point \((0,0)\).

Step4: Construct the Graph

• For \(x\lt1\): The function should approach \(\infty\) as \(x\to1^{-}\) and pass through \((0,0)\). We can consider a function like \(y=\frac{-2x(x - 1)+0}{x - 1}\) (but a simpler way is to think of a curve that comes from the horizontal asymptote \(y = - 2\) (as \(x\to-\infty\)), goes through \((0,0)\) and then goes to \(\infty\) as \(x\to1^{-}\).
• For \(x>1\): The function should approach \(-\infty\) as \(x\to1^{+}\) and approach \(y=-2\) as \(x\to\infty\).
• At \(x = 0\), the function has a value of \(0\). At \(x = 1\), the function has a vertical asymptote.

To sketch the graph:

1. Draw the horizontal line \(y=-2\) (horizontal asymptote).
2. Draw the vertical line \(x = 1\) (vertical asymptote).
3. Plot the point \((0,0)\).
4. For \(x\lt1\), draw a curve that starts from the left - hand side of the horizontal asymptote (\(y=-2\) as \(x\to-\infty\)), passes through \((0,0)\) and then goes up towards the vertical asymptote \(x = 1\) (going to \(\infty\) as \(x\to1^{-}\)).
5. For \(x>1\), draw a curve that comes from the right - hand side of the vertical asymptote (going to \(-\infty\) as \(x\to1^{+}\)) and then approaches the horizontal asymptote \(y = - 2\) as \(x\to\infty\).

(Note: A possible function that satisfies these conditions is \(f(x)=\frac{-2x}{x - 1}\). Let's check:

• \(f(0)=\frac{0}{-1}=0\)
• \(\lim_{x\to1^{-}}\frac{-2x}{x - 1}=\infty\) (since as \(x\to1^{-}\), \(x - 1\to0^{-}\) and \(-2x\to - 2\), so \(\frac{-2x}{x - 1}\to\infty\))
• \(\lim_{x\to1^{+}}\frac{-2x}{x - 1}=-\infty\) (since as \(x\to1^{+}\), \(x - 1\to0^{+}\) and \(-2x\to - 2\), so \(\frac{-2x}{x - 1}\to-\infty\))
• \(\lim_{x\to\pm\infty}\frac{-2x}{x - 1}=\lim_{x\to\pm\infty}\frac{-2}{1-\frac{1}{x}}=-2\) (dividing numerator and denominator by \(x\))

Answer:

The graph has a horizontal asymptote \(y = - 2\), a vertical asymptote \(x = 1\), passes through \((0,0)\), approaches \(\infty\) as \(x\to1^{-}\), approaches \(-\infty\) as \(x\to1^{+}\), and approaches \(-2\) as \(x\to\pm\infty\). A possible function is \(f(x)=\frac{-2x}{x - 1}\) and its graph can be sketched following the steps above.