Sketch a rational function subject to the given conditions. Horizontal asymptote: \( y=2 \) Vertical asymptote: \( x=1 \) \( y \)-intercept: \( (0,3) \) \( x \)-intercept: \( \left(\frac{3}{2}, 0\right) \)

Consider a rational function of the form \[ f(x)=\frac{p(x)}{q(x)} \] where we choose \( p(x) \) and \( q(x) \) such that all conditions are satisfied. **Step 1. Horizontal asymptote: \( y=2 \)** For the horizontal asymptote to be \( y=2 \), the degrees of the numerator and denominator must be the same and the ratio of their leading coefficients must be 2. A natural choice is to let both the numerator and denominator be linear. Let \[ p(x)=2x+b \quad \text{and} \quad q(x)=x-1. \] The leading coefficient of the numerator is 2 and that of the denominator is 1, so \[ \lim_{x \to \infty} f(x)=\frac{2}{1}=2. \] This satisfies the horizontal asymptote condition. **Step 2. Vertical asymptote: \( x=1 \)** For \( x=1 \) to be a vertical asymptote, the denominator must be zero at \( x=1 \) (and not cancel with the numerator). With \[ q(x)=x-1, \] we have \( q(1)=0 \). Therefore, the function has a vertical asymptote at \( x=1 \). **Step 3. \( y \)-intercept: \( (0,3) \)** The \( y \)-intercept is given when \( x=0 \): \[ f(0)=\frac{2(0)+b}{0-1}=\frac{b}{-1}=-b. \] We need \[ -b=3 \quad \Rightarrow \quad b=-3. \] Thus, the numerator becomes \[ p(x)=2x-3. \] **Step 4. \( x \)-intercept: \( \left(\frac{3}{2},0\right) \)** The \( x \)-intercept is found by setting the numerator equal to zero: \[ 2x-3=0 \quad \Rightarrow \quad 2x=3 \quad \Rightarrow \quad x=\frac{3}{2}. \] This matches the given \( x \)-intercept. **Final Function** The function that meets all the conditions is \[ f(x)=\frac{2x-3}{x-1}. \]