Question

find the vertex of f(x)=4x^2+8x-2

Answer 1

Answer:

$f(x) =4x^2 + 8x -2 \\f(x) = 4(x^2 + 2x -\frac{1}{2} )\\\\f(x) = 4(x^2 + 2x +1 - 1 -\frac{1}{2} )\\f(x) = 4((x+1)^2 - 1 -\frac{1}{2} ))\\f(x) = 4((x+1)^2 -\frac{3}{2} ))\\f(x) = 4(x+1)^2 -6$

the vertices (-1, -6)