Sketch a rational function subject to the given conditions. Horizontal asymptote: \( y=3 \) Vertical asymptotes: \( x=-1 \) and \( x=1 \) \( y \)-intercept: \( (0,0) \) \( x \)-intercept: \( (0,0) \) Symmetric to the \( y \)-axis Passes through the point \( (3,6) \)

To construct a rational function that meets all your conditions, we can start by considering the general form of a rational function: \[ f(x) = \frac{p(x)}{q(x)} \] Where \( p(x) \) is the numerator and \( q(x) \) is the denominator. Given that there are vertical asymptotes at \( x = -1 \) and \( x = 1 \), our denominator will have factors of \( (x + 1)(x - 1) \). This can be expressed as: \[ q(x) = (x + 1)(x - 1) = x^2 - 1 \] Next, since the horizontal asymptote is \( y = 3 \), the degrees of the numerator and denominator must be equal. Therefore, we need \( p(x) \) such that its leading coefficient is 3. Given that the function is symmetric about the \( y \)-axis, the numerator must be an even function; hence we can write: \[ p(x) = 3x^2 \] Now we have the function: \[ f(x) = \frac{3x^2}{x^2 - 1} \] To check if it satisfies the remaining conditions, we note that: - \( f(0) = \frac{3(0)^2}{0^2 - 1} = 0 \) confirms the \( y \)-intercept and \( x \)-intercept at \( (0,0) \). - Substituting \( x = 3 \): \( f(3) = \frac{3(3)^2}{(3)^2 - 1} = \frac{27}{8} = 3.375 \). We still need to adjust to pass through the point \( (3,6) \). To ensure the function passes through \( (3,6) \), we can introduce a constant multiplier \( k \): \[ f(x) = k \cdot \frac{3x^2}{x^2 - 1} \] By substituting \( x = 3 \) and setting \( f(3) = 6 \): \[ 6 = k \cdot \frac{27}{8} \] \[ k = 6 \cdot \frac{8}{27} = \frac{48}{27} = \frac{16}{9} \] Thus, the complete function becomes: \[ f(x) = \frac{16}{9} \cdot \frac{3x^2}{x^2 - 1} = \frac{48x^2}{9(x^2 - 1)} = \frac{48x^2}{9(x^2 - 1)} \] This function satisfies all your conditions! Now you can sketch \( f(x) \) with these properties in mind, highlighting the asymptotes and intercepts. Happy sketching!