Question

What is the axis of symmetry of h(x)=5x^+40x+64

Answer 1

$\text{The axis of symmetry of}\ f(x)=ax^2+bx+c\ is\ x=\dfrac{-b}{2a}.\\\\\text{We have:}\ h(x)=5x^2+40x+64\\\\\text{substitute:}\ x=\dfrac{-40}{2\cdot5}=\dfrac{-40}{10}=-4.\\\\Answer:\ \boxed{x=-4}$