Question

Y=-x^2-10x+24 what is the u value of the vertex

Answer 1

`\large\boxed{The\ vertex:\ (-5,\ 49)}`

Explanation:

`f(x)=ax^2+bx+c\\\\\text{The vertex}\ (h,\ k),\ \text{where}\ h=(-b)/(2a)\ \text{and}\ k=f(h).\\\\\text{We have:}\ y=-x^2-10x+24\to f(x)=-x^2-10x+24\\\\a=-1,\ b=-10,\ c=24.\\\\\text{Substitute:}\\\\h=(-(-10))/(2(-1))=(10)/(-2)=-5\\\\k=f(-5)=-(-5)^2-10(-5)+24=-25+50+24=49`