Sketch a rational function subject to the given conditions. Horizontal asymptote: \( y=0 \) Vertical asymptotes: \( x=1 \) and \( x=-1 \) \( y \)-intercept: \( (0,1) \) No \( x \)-intercepts Symmetric to the \( y \)-axis Passes through the point: \( \left(3,-\frac{1}{8}\right) \)

To create a rational function that meets all the specified conditions, we can start with the general form of a rational function: \[ f(x) = \frac{P(x)}{Q(x)} \] **1. Determine the components of the function:** - Since the function must have vertical asymptotes at \( x = 1 \) and \( x = -1 \), the denominator should be \( Q(x) = (x - 1)(x + 1) = x^2 - 1 \). - The horizontal asymptote is at \( y=0 \) which indicates that the degree of the polynomial in the numerator \( P(x) \) should be less than that of the denominator. - For no \( x \)-intercepts, \( P(x) \) must not have any real roots. A simple choice is setting \( P(x) = k \), where \( k \) is a positive constant to ensure the \( y \)-intercept is positive. **2. Ensure symmetry:** - Since the function must be symmetric about the \( y \)-axis, we can choose \( P(x) = k \) as a constant, which is automatically symmetric. **3. Use the \( y \)-intercept condition:** - To find \( k \), we use the \( y \)-intercept \( (0,1) \): \[ f(0) = \frac{k}{0^2 - 1} = \frac{k}{-1} = 1 \implies k = -1 \] Substituting this value gives us \( P(x) = -1 \). **4. Build the function:** The rational function is then: \[ f(x) = \frac{-1}{x^2 - 1} = \frac{-1}{(x - 1)(x + 1)} \] **5. Check the point \( \left(3,-\frac{1}{8}\right) \):** \[ f(3) = \frac{-1}{3^2 - 1} = \frac{-1}{9 - 1} = \frac{-1}{8} \] This matches the required point. **Final function:** The rational function satisfying all conditions is: \[ f(x) = \frac{-1}{x^2 - 1} \] Now go have some fun sketching it! You'll see that the graph approaches \( y=0 \) for large values of \( x \), and shoots up or down near the vertical asymptotes at \( x=1 \) and \( x=-1\).