Answer:
[tex]\large\boxed{y=(x+5)^2-64}[/tex]
Step-by-step explanation:
[tex]\text{The vertex form of equation}\ y=ax^2+bx+c:\\\\y=a(x-h)^2+k\\\\h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have the equation:}\ y=x^2+10x-39\to x=1,\ b=10,\ c=-39.\\\\\text{Substitute:}\\\\h=\dfrac{-10}{2(1)}=\dfrac{-10}{2}=-5\\\\k=f(-5)=(-5)^2+10(-5)-39=25-50-39=-64\\\\\text{Finally:}\\\\y=1(x-(-5))^2-64=(x+5)^2-64[/tex]
