Next question You can retry this question Write an equation for a rational function with: Vertical asymptotes at \( x=-1 \) and \( x=2 \) \( x \) intercepts at \( x=-3 \) and \( x=5 \) \( y \) intercept at 6 \( y= \) Ouestion Hell.

To create a rational function that has vertical asymptotes at \(x = -1\) and \(x = 2\), we can start with the denominator: \((x + 1)(x - 2)\). For the x-intercepts at \(x = -3\) and \(x = 5\), we will use the numerator: \((x + 3)(x - 5)\). Thus, the function can be expressed as: \[ y = \frac{(x + 3)(x - 5)}{(x + 1)(x - 2)} \] Now we need to adjust the function so that the y-intercept equals 6. To find the constant \(k\), we substitute \(x = 0\): \[ y(0) = k \cdot \frac{(0 + 3)(0 - 5)}{(0 + 1)(0 - 2)} = k \cdot \frac{3(-5)}{1(-2)} = k \cdot \frac{-15}{-2} = \frac{15k}{2} \] Setting this equal to 6 gives us: \[ \frac{15k}{2} = 6 \] Solving for \(k\) leads to \(k = \frac{6 \cdot 2}{15} = \frac{12}{15} = \frac{4}{5}\). Putting it all together, the rational function is: \[ y = \frac{4/5 \cdot (x + 3)(x - 5)}{(x + 1)(x - 2)} \] Now you’ve got a function that meets all of the criteria you specified!