Question

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

Answer 1

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:
\[ f(x) = \frac{(x + 1)(x + 2)}{(2x + 3)(x - 2)} \]

2. **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 \( (0, -\frac{1}{3}) \).

3. **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.

4. **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:
\[ y = \frac{1}{2} \]

5. **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.