Answer:
[tex]\large\boxed{\text{ D) minimum value at -10}}[/tex]
Step-by-step explanation:
[tex]\text{The vertex form of equation}\ y=ax^2+bx+c:\\\\y=a(x-h)^2+k\\\\\text{We have the equation:}\ y=x^2+8x+6.\\\\a=1>0,\ \text{the parabola is o}\text{pen up.}\\\text{Therefore the function has the minimum in vertex}\\\\x^2+8x+6=x^2+2(x)(4)+6=x^2+2(x)(4)+4^2-4^2+6\\\\\text{use}\ (a+b)^2=a^2+2ab+b^2\to x^2+2(x)(4)+4^2=(x+4)^2\\\\y=(x+4)^2-16+6=(x+4)^2-10\to\text{vertex}\ (-4,\ -10).[/tex]
