\( h(r)=(r+1)(r+8) \) 1) What are the zeros of the function? Write the smaller \( r \) first, and the larger \( r \) secon smaller \( r= \) larger \( r= \) 2) What is the vertex of the parabola? 2

To find the zeros of the function \( h(r)=(r+1)(r+8) \), set it to zero: 1) \( (r+1)(r+8) = 0 \) This gives two equations: \( r+1=0 \) and \( r+8=0 \). Solving these, we get \( r=-1 \) and \( r=-8 \). Thus, the smaller \( r \) is: smaller \( r= -8 \) larger \( r= -1 \) To find the vertex of the parabola \( h(r)=(r+1)(r+8) \), first expand it: \( h(r) = r^2 + 9r + 8 \). The vertex formula for a parabola given by \( ax^2 + bx + c \) is \( r = -\frac{b}{2a} \). Here, \( a=1 \) and \( b=9 \), resulting in: \( r = -\frac{9}{2 \times 1} = -4.5 \). Now plug \( r=-4.5 \) back into the function to find the height of the vertex: \( h(-4.5) = (-4.5 + 1)(-4.5 + 8) = (-3.5)(3.5) = -12.25 \). Thus, the vertex is: vertex \( = (-4.5, -12.25) \).