Replace the expression -u-5 for z into the rule of correspondence of h to find h(-u-5):
[tex]\begin{gathered} h(z)=z^2 \\ \Rightarrow h(-u-5)=(-u-5)^2 \end{gathered}[/tex]
Expand the square binomial using the formula (a+b)^2=a^2+2ab+b^2, with a=-u and b=-5:
[tex]\begin{gathered} (-u-5)^2=(-u)^2+2(-u)(-5)+(-5)^2 \\ =u^2+10u+25 \end{gathered}[/tex]
Therefore:
[tex]h(-u-5)=u^2+10u+25[/tex]
