To analyze the rational function \( f(x) = \frac{x^2 + 3x + 2}{2x^2 - x - 2} \), we should first determine its characteristics, such as vertical and horizontal asymptotes.
- Factor the numerator and denominator:
- The numerator: \( x^2 + 3x + 2 = (x+1)(x+2) \)
- The denominator: \( 2x^2 - x - 2 \) To factor this, we can look for factors of \( 2 \times -2 = -4 \) that sum to -1. These factors are -2 and +2. Thus, the denominator can be factored by using synthetic division or decomposition, resulting in \( 2x^2 - 4x + 3x - 2 = (2x + 3)(x - 1) \).
So, we have: \[ f(x) = \frac{(x+1)(x+2)}{(2x+3)(x-1)} \]
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Determine vertical asymptotes:
- Vertical asymptotes occur where the denominator equals zero, i.e., \( 2x + 3 = 0 \) or \( x - 1 = 0 \).
- Solving these gives:
- \( x = -\frac{3}{2} \) (vertical asymptote)
- \( x = 1 \) (vertical asymptote)
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Determine horizontal asymptotes:
- The degrees of the numerator and the denominator are the same. The horizontal asymptote is the ratio of the leading coefficients.
- The leading coefficient of the numerator is 1 and the leading coefficient of the denominator is 2:
- Thus, the horizontal asymptote is \( y = \frac{1}{2} \).
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Intercepts:
- The x-intercepts occur when the numerator is zero, i.e., \( x + 1 = 0 \) or \( x + 2 = 0 \) gives \( x = -1 \) and \( x = -2 \).
- The y-intercept occurs when \( x = 0 \): \( f(0) = \frac{(0 + 1)(0 + 2)}{(2(0) + 3)(0 - 1)} = \frac{2}{-3} = -\frac{2}{3} \).
Now, we can match these characteristics with the provided graph descriptions.
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Vertical Asymptotes: Lines at \( x = -1.5 \) and \( x = 1 \).
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Horizontal Asymptote: Line at \( y = \frac{1}{2} \).
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X-intercepts: Points at \( (-1, 0) \) and \( (-2, 0) \).
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Y-intercept: Point at \( (0, -\frac{2}{3}) \).
Given these characteristics, the right graph will have the appropriate behavior: the left branch should start below the line \( y = \frac{1}{2} \) and approach \( x = 1 \) from below. The right branch should start above \( y = \frac{1}{2} \) and curve downwards.
The first option matches that description with the horizontal and vertical asymptotes drawn and the curve behavior as described.
Therefore, the answer is the first graph option.