Question

What is the vertex of f(x) = x^2 + 10x + 16?

Answer 1

`f(x)=x^2+10x+16\\\\V(-(b)/(2a),-(\Delta)/(4a))-vertex\\\\-(b)/(2a)=-(10)/(2\cdot1)=-5\\\\\Delta=b^2-4ac\\\Delta=10^2-4\cdot1\cdot16=100-64=36\\\\-(\Delta)/(4a)=-(36)/(4\cdot1)=-9\\\\V(-5,-9)`