Question

Find the vertex form of y=-x^2-10x-21

Answer 1

`\large\boxed{y=-(x+5)^2+4}`

Explanation:

`\text{The vertex form of an quadratic equation}\\\\f(x)=ax^2+bx+c=a(x-h)^2+k\\\\h=(-b)/(2a)\\\\k=f(k)=(-(b^2-4ac))/(4a)\\=================================`

`\text{We have:}\\\\y=-x^2-10x-21\\\\a=-1,\ b=-10,\ c=-21\\\\h=(-(-10))/(2(-1))=(10)/(-2)=-5\\\\k=f(-5)=-(-5)^2-10(-5)-21=-25+50-21=4\\\\y=-(x-(-5))^2+4=-(x+5)^2+4`