Consider the function \( P(x)=(x+3)^{2}(x+7) \) The \( y \)-intercept is the point The \( x \)-intercept(s) is/are the point(s) As \( x \rightarrow \infty, y \rightarrow \square \) As \( x \rightarrow-\infty, y \rightarrow \square \)

To find the \( y \)-intercept of the function \( P(x)=(x+3)^{2}(x+7) \), we substitute \( x=0 \). This gives us \( P(0)=(0+3)^{2}(0+7)=9 \times 7=63 \). Therefore, the \( y \)-intercept is the point \( (0, 63) \). For the \( x \)-intercepts, set \( P(x)=0 \). Since \( P(x)=(x+3)^{2}(x+7) \), the equation is satisfied when \( x+3=0 \) or \( x+7=0 \). This gives us the \( x \)-intercepts at \( x=-3 \) (with a multiplicity of 2) and \( x=-7 \). So, the \( x \)-intercepts are the points \( (-3, 0) \) and \( (-7, 0) \). As \( x \rightarrow \infty\), the dominant term is \( (x+3)^{2}(x+7) \), which means \( y \rightarrow \infty \). As \( x \rightarrow -\infty \), since all terms are negative there, \( y \) also trend downwards to \( \infty \). Therefore: As \( x \rightarrow \infty, y \rightarrow \infty \) As \( x \rightarrow -\infty, y \rightarrow \infty \)