# Step-by-Step Solution Let's use the key features from your description and the graph: - **Vertical asymptotes:** \(x = -1\), \(x = 2\) - **Horizontal asymptote:** \(y = 0\) - **y-intercept:** \((0, 2)\) - **x-intercept:** \((1, 0)\) ## 1. **Form of the Rational Function** Since there are vertical asymptotes at \(x = -1\) and \(x = 2\), the denominator must be zero at those points: \[ (x + 1)(x - 2) \] Let the function be: \[ y = \frac{A(x - h)}{(x + 1)(x - 2)} \] where \(A\) and \(h\) are constants to be found. ## 2. **x-intercept at \(x=1\)** Set \(y = 0\) and \(x=1\): \[ 0 = \frac{A(1 - h)}{(1 + 1)(1 - 2)} \] Numerator must be zero: \[ 1 - h = 0 \implies h = 1 \] So now: \[ y = \frac{A(x - 1)}{(x + 1)(x - 2)} \] ## 3. **y-intercept at \(x=0\): \((0, 2)\)** Plug \(x=0\), \(y=2\): \[ 2 = \frac{A(0 - 1)}{(0 + 1)(0 - 2)} \] \[ 2 = \frac{-A}{1 \times -2} = \frac{-A}{-2} = \frac{A}{2} \] \[ A = 4 \] ## 4. **Final Equation** Combine all found values: \[ \boxed{y = \frac{4(x-1)}{(x+1)(x-2)}} \] ## 5. **Check: Horizontal Asymptote \(y=0\)** As \(x \to \infty\), numerator and denominator are both degree 1 and 2, so \(y \to 0\) as required. --- ## **Final Answer** \[ \boxed{y = \frac{4(x-1)}{(x+1)(x-2)}} \] **Demonstrating two equal signs in one line:** \[ y = \frac{4(x-1)}{(x+1)(x-2)} = \frac{4x-4}{x^2-x-2} \]