Given the function \( P(x)=x(x-7)(x+2) \), find its \( y \)-intercept: its \( x \)-intercepts: When \( x \rightarrow \infty, y \rightarrow \) ? When \( x \rightarrow-\infty, y \rightarrow \) ?

To find the \( y \)-intercept of the function \( P(x)=x(x-7)(x+2) \), set \( x = 0 \). Plugging this in gives \( P(0) = 0(0-7)(0+2) = 0 \). So, the \( y \)-intercept is at the point \( (0, 0) \). For the \( x \)-intercepts, set \( P(x) = 0 \). The equation \( x(x - 7)(x + 2) = 0 \) gives us the intercepts at \( x = 0 \), \( x = 7 \), and \( x = -2 \). Thus, the \( x \)-intercepts are at \( (0, 0) \), \( (7, 0) \), and \( (-2, 0) \). As \( x \rightarrow \infty \), \( P(x) \) behaves like \( x^3 \) since the leading term dominates, leading us to conclude that \( y \rightarrow \infty \). Conversely, as \( x \rightarrow -\infty \), the leading term \( x^3 \) will result in \( y \rightarrow -\infty \) as well, since the cube of a negative number remains negative!