Respuesta :
For the zeros of the function, we have to solve h(r)=0, therefore:
[tex]\begin{gathered} h(r)=(r+1)(r+8) \\ h(r)=0 \\ \Rightarrow(r+1)(r+8)=0 \\ \Rightarrow r=-1\text{ or } \\ r=-8 \end{gathered}[/tex]then, the smaller r is -8 and the larger is -1.
Now, to find the vertex of the parabola, we can find the x-coordinate of the vertex from the general rule:
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ \text{ x-coordinate: -b/2a} \end{gathered}[/tex]In this case, we have the following:
[tex]\begin{gathered} h(r)=(r+1)(r+8)=r^2+8r+r+8=r^2+9r+8 \\ \Rightarrow a=1,b=9 \\ \Rightarrow-\frac{b}{2a}=-\frac{9}{2(1)}=-\frac{9}{2} \end{gathered}[/tex]now that we have the x-coordinate of the vertex, we just evaluate the function on that point to find the y-coordinate of the vertex:
[tex]h(-\frac{9}{2})=(-\frac{9}{2}+1)(-\frac{9}{2}+8)=(-\frac{7}{2})(\frac{7}{2})=-\frac{49}{4}[/tex]therefore, the vertex of the parabola is the point (-9/2,-49/4)
