Grade 11 Mathematics Worksheet: Functions - Hyperbola - Free Printable

Problem Overview:

The task is to match the given hyperbolic functions with their corresponding graphs. The functions are provided in the form of equations, and we need to analyze their characteristics (such as asymptotes, intercepts, and general shape) to identify which graph corresponds to each function.

Given Functions:

1. \( y = \frac{3}{x-1} - 1 \)
2. \( y = \frac{1}{x-2} \)
3. \( y = -\frac{3}{x+2} \)
4. \( y = -\frac{2}{x} + 1 \)

Graphs:

There are four graphs labeled A, B, C, and D. We will analyze each function and match it with the appropriate graph.

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Step-by-Step Analysis:

#### Function 1.1: \( y = \frac{3}{x-1} - 1 \)

1. Vertical Asymptote: The denominator \( x-1 \) becomes zero when \( x = 1 \). Therefore, the vertical asymptote is \( x = 1 \).
2. Horizontal Asymptote: As \( x \to \pm\infty \), \( \frac{3}{x-1} \to 0 \), so \( y \to -1 \). Thus, the horizontal asymptote is \( y = -1 \).
3. Behavior: The function \( \frac{3}{x-1} \) is positive for \( x > 1 \) and negative for \( x < 1 \). Subtracting 1 shifts the entire graph down by 1 unit.
4. Graph Match: The graph should have a vertical asymptote at \( x = 1 \) and a horizontal asymptote at \( y = -1 \). The branches of the hyperbola should be in the regions where \( x > 1 \) and \( x < 1 \).

- Matching Graph: Graph D.

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#### Function 1.2: \( y = \frac{1}{x-2} \)

1. Vertical Asymptote: The denominator \( x-2 \) becomes zero when \( x = 2 \). Therefore, the vertical asymptote is \( x = 2 \).
2. Horizontal Asymptote: As \( x \to \pm\infty \), \( \frac{1}{x-2} \to 0 \). Thus, the horizontal asymptote is \( y = 0 \).
3. Behavior: The function \( \frac{1}{x-2} \) is positive for \( x > 2 \) and negative for \( x < 2 \).
4. Graph Match: The graph should have a vertical asymptote at \( x = 2 \) and a horizontal asymptote at \( y = 0 \). The branches of the hyperbola should be in the regions where \( x > 2 \) and \( x < 2 \).

- Matching Graph: Graph B.

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#### Function 1.3: \( y = -\frac{3}{x+2} \)

1. Vertical Asymptote: The denominator \( x+2 \) becomes zero when \( x = -2 \). Therefore, the vertical asymptote is \( x = -2 \).
2. Horizontal Asymptote: As \( x \to \pm\infty \), \( -\frac{3}{x+2} \to 0 \). Thus, the horizontal asymptote is \( y = 0 \).
3. Behavior: The negative sign in front of the fraction flips the branches of the hyperbola. The function is negative for \( x > -2 \) and positive for \( x < -2 \).
4. Graph Match: The graph should have a vertical asymptote at \( x = -2 \) and a horizontal asymptote at \( y = 0 \). The branches of the hyperbola should be in the regions where \( x > -2 \) and \( x < -2 \), but flipped compared to a standard hyperbola.

- Matching Graph: Graph A.

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#### Function 1.4: \( y = -\frac{2}{x} + 1 \)

1. Vertical Asymptote: The denominator \( x \) becomes zero when \( x = 0 \). Therefore, the vertical asymptote is \( x = 0 \).
2. Horizontal Asymptote: As \( x \to \pm\infty \), \( -\frac{2}{x} \to 0 \), so \( y \to 1 \). Thus, the horizontal asymptote is \( y = 1 \).
3. Behavior: The negative sign in front of the fraction flips the branches of the hyperbola. Adding 1 shifts the entire graph up by 1 unit. The function is negative for \( x > 0 \) and positive for \( x < 0 \).
4. Graph Match: The graph should have a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 1 \). The branches of the hyperbola should be in the regions where \( x > 0 \) and \( x < 0 \), but flipped compared to a standard hyperbola.

- Matching Graph: Graph C.

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Final Matches:

1. \( y = \frac{3}{x-1} - 1 \) → Graph D
2. \( y = \frac{1}{x-2} \) → Graph B
3. \( y = -\frac{3}{x+2} \) → Graph A
4. \( y = -\frac{2}{x} + 1 \) → Graph C

Final Answer:

\[
\boxed{1.1 \to D, 1.2 \to B, 1.3 \to A, 1.4 \to C}
\]