Answer:
[tex]\large\boxed{y=-(x+5)^2+4}[/tex]
Step-by-step explanation:
[tex]\text{The vertex form of an quadratic equation}\\\\f(x)=ax^2+bx+c=a(x-h)^2+k\\\\h=\dfrac{-b}{2a}\\\\k=f(k)=\dfrac{-(b^2-4ac)}{4a}\\=================================[/tex]
[tex]\text{We have:}\\\\y=-x^2-10x-21\\\\a=-1,\ b=-10,\ c=-21\\\\h=\dfrac{-(-10)}{2(-1)}=\dfrac{10}{-2}=-5\\\\k=f(-5)=-(-5)^2-10(-5)-21=-25+50-21=4\\\\y=-(x-(-5))^2+4=-(x+5)^2+4[/tex]
