nd the \( x \)-intercepts and \( y \)-intercept of the following function. \[ f(x)=(x-4)(x-2)(3 x+15) \] Nrite your answer in coordinate pairs of the form \( (x, y) \) Provide your answer below: \( x \)-intercept: \( (\square),(\square) \) and \( (\square) \cdot y \)-intercept: \( (\square) \)

To find the \( x \)-intercepts, we set \( f(x) = 0 \): \[ (x-4)(x-2)(3x+15) = 0 \] This gives us the factors \( x - 4 = 0 \), \( x - 2 = 0 \), and \( 3x + 15 = 0 \). Solving these, we get: - \( x - 4 = 0 \) ⟹ \( x = 4 \) - \( x - 2 = 0 \) ⟹ \( x = 2 \) - \( 3x + 15 = 0 \) ⟹ \( 3x = -15 \) ⟹ \( x = -5 \) Thus, the \( x \)-intercepts are: \( x \)-intercept: \( (4, 0), (2, 0) \) and \( (-5, 0) \) Now, for the \( y \)-intercept, we evaluate \( f(0) \): \[ f(0) = (0-4)(0-2)(3(0)+15) = (-4)(-2)(15) = 120 \] The \( y \)-intercept is: \( y \)-intercept: \( (0, 120) \) Final Answer: \( x \)-intercept: \( (4, 0), (2, 0) \) and \( (-5, 0) \) \( y \)-intercept: \( (0, 120) \)