Question

1. construct a graph of a function that satisfies the following properties: $lim_{x \to \infty} h(x) = 0$, $lim_{x \to -\infty} h(x) = 0$, $lim_{x \to 2^+} h(x) = \infty$, $lim_{x \to 2^-} h(x) = -\infty$, and $h(0) = -1$.

Explanation:

Step1: Analyze Asymptotes and Intercept

We need a function with horizontal asymptotes \( y = 0 \) (since \( \lim_{x
ightarrow\infty}h(x)=0 \) and \( \lim_{x
ightarrow-\infty}h(x)=0 \)), a vertical asymptote at \( x = 2 \) with \( \lim_{x
ightarrow2^{+}}h(x)=\infty \) and \( \lim_{x
ightarrow2^{-}}h(x)=-\infty \), and \( h(0)=-1 \). A rational function of the form \( h(x)=\frac{a}{(x - 2)}+b \) can be considered. Since horizontal asymptote is \( y = 0 \), \( b = 0 \). Now, use \( h(0)=-1 \): \( h(0)=\frac{a}{(0 - 2)}=-1\Rightarrow\frac{a}{-2}=-1\Rightarrow a = 2 \). So the function is \( h(x)=\frac{2}{x - 2} \).

Step2: Verify Properties

• Horizontal Asymptotes: For \( \lim_{x

ightarrow\pm\infty}\frac{2}{x - 2} \), as \( x
ightarrow\pm\infty \), \( x - 2
ightarrow\pm\infty \), so \( \frac{2}{x - 2}
ightarrow0 \), which matches \( \lim_{x
ightarrow\infty}h(x)=0 \) and \( \lim_{x
ightarrow-\infty}h(x)=0 \).

• Vertical Asymptote: The denominator \( x - 2 = 0 \) when \( x = 2 \). For \( \lim_{x

ightarrow2^{+}}\frac{2}{x - 2} \), as \( x
ightarrow2^{+} \), \( x - 2
ightarrow0^{+} \), so \( \frac{2}{x - 2}
ightarrow\infty \). For \( \lim_{x
ightarrow2^{-}}\frac{2}{x - 2} \), as \( x
ightarrow2^{-} \), \( x - 2
ightarrow0^{-} \), so \( \frac{2}{x - 2}
ightarrow-\infty \), which matches the given limits.

• Value at \( x = 0 \): \( h(0)=\frac{2}{0 - 2}=\frac{2}{-2}=-1 \), which matches \( h(0)=-1 \).

To graph \( h(x)=\frac{2}{x - 2} \):

1. Vertical Asymptote: Draw a dashed line at \( x = 2 \).
2. Horizontal Asymptote: Draw a dashed line at \( y = 0 \).
3. Points: Plot \( (0,-1) \). For \( x>2 \) (e.g., \( x = 3 \)), \( h(3)=\frac{2}{3 - 2}=2 \), so the point \( (3,2) \) is on the right - hand branch. For \( x<2 \) (e.g., \( x = 1 \)), \( h(1)=\frac{2}{1 - 2}=-2 \), so the point \( (1,-2) \) is on the left - hand branch.
4. Shape: The right - hand branch (for \( x>2 \)) approaches \( x = 2 \) from the right going up to \( \infty \) and approaches \( y = 0 \) as \( x

ightarrow\infty \). The left - hand branch (for \( x<2 \)) approaches \( x = 2 \) from the left going down to \( -\infty \) and approaches \( y = 0 \) as \( x
ightarrow-\infty \).

Answer:

A function that satisfies the properties is \( h(x)=\frac{2}{x - 2} \). The graph has a vertical asymptote at \( x = 2 \), a horizontal asymptote at \( y = 0 \), passes through the point \( (0,-1) \), with the right - hand branch (for \( x>2 \)) going from \( (2,\infty) \) towards \( (3,2) \) and approaching \( y = 0 \) as \( x
ightarrow\infty \), and the left - hand branch (for \( x<2 \)) going from \( (-\infty,2) \) towards \( (1,-2) \) and approaching \( y = 0 \) as \( x
ightarrow-\infty \).