Part A:
Given function is
[tex]y=-3x^2-6x-5[/tex]
While comparing the above function with the square of the function
[tex]y=a(x-h)^2+k[/tex]
then, we have
[tex]\begin{gathered} y=-3x^2-6x-5=a(x-h)^2+k \\ =-3(x-h)^2+k \\ =-3(x^2-2hx+h^2)+k \\ =-3x^2+6hx-3h^2+k \\ =-3x^2+6hx+(k-3h^2) \end{gathered}[/tex]
while comparing the above equations,
[tex]\begin{gathered} y=-3x^2-6x-5=-3x^2+6hx+(k-3h^2) \\ \therefore a=-3\text{ } \\ \Rightarrow-6=6h \\ \therefore h=-1 \\ \Rightarrow-5=k-3h^2 \\ -5+3(-1)^2=k \\ k=-5+3 \\ \therefore k=-2 \end{gathered}[/tex]
therefore, the square of the function will be
[tex]\begin{gathered} \text{substituting a=-3 , h=-1 and k=-2 in Square of the function} \\ y=a(x-h)^2+k \\ \therefore y=-3(x+1)^2-2 \end{gathered}[/tex]
