Which graph shows the rational function f(x)=x2+3x+2x2−x−2

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1 answer

To analyze the rational function \( f(x) = \frac{x^2 + 3x + 2}{2x^2 - x - 2} \), we should first simplify it if possible, and then determine characteristics such as intercepts, asymptotes, and behavior at infinity to visualize what the graph should look like.

  1. Factoring the numerator and denominator:
    • For the numerator \( x^2 + 3x + 2 \), we can factor it: \[ x^2 + 3x + 2 = (x + 1)(x + 2) \]
    • For the denominator \( 2x^2 - x - 2 \), we need to factor it as well: \[ 2x^2 - x - 2 = (2x + 3)(x - 2) \]

Thus, the function can be rewritten as: \[ f(x) = \frac{(x + 1)(x + 2)}{(2x + 3)(x - 2)} \]

  1. Finding intercepts:

    • x-intercepts: Set the numerator equal to zero: \[ (x + 1)(x + 2) = 0 \] This gives us \( x = -1 \) and \( x = -2 \) as x-intercepts.
    • y-intercept: Set \( x = 0 \): \[ f(0) = \frac{(0 + 1)(0 + 2)}{(2(0) + 3)(0 - 2)} = \frac{1 \cdot 2}{3 \cdot (-2)} = \frac{2}{-6} = -\frac{1}{3} \] So the y-intercept is \( (0, -\frac{1}{3}) \).
  2. Vertical asymptotes: Set the denominator equal to zero: \[ (2x + 3)(x - 2) = 0 \] This gives \( x = -\frac{3}{2} \) and \( x = 2 \) as vertical asymptotes.

  3. Horizontal asymptote: Since the degrees of the numerator and denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients: \[ y = \frac{1}{2} \]

  4. End behavior: As \( x \to \pm \infty \), \( f(x) \to \frac{1}{2} \).

With this information, we can summarize:

  • The x-intercepts are \( -1 \) and \( -2 \).
  • The y-intercept is \( (0, -\frac{1}{3}) \).
  • Vertical asymptotes at \( x = -\frac{3}{2} \) and \( x = 2 \).
  • A horizontal asymptote at \( y = \frac{1}{2} \).

If you have specific graphs presented, look for one that includes the intercepts at \( (-1, 0) \) and \( (-2, 0) \), has vertical asymptotes at \( x = -\frac{3}{2} \) and \( x = 2 \), and approaches \( y = \frac{1}{2} \) as \( x \) approaches infinity.