Let fx 2x24x1 Put f into standard form by completing the sq

Let f(x) = 2x^2-4x-1. Put f into standard form by completing the square and then identify: y = 2(x^2 - 2x)_-1 y + 1 = 2 (x^2 - 2x) Find the vertex y - (-3) = 2(x - 1)^2 Find the x-symmetry (vertical line) Find zeros or x-intercept y-intercept: Graph parabola to find Domain: Range:

Solution

f(x) = 2x^2 -4x -1

Let y = 2x^2 -4x -1

y = 2( x^2 -2x +1 -1) -1

y = 2(x-1)^2 -3

Standard vertex form of quadratic equation: y = a( x- h)^2 +k

where ( h, k) is vertex

a) So, comparing with our equation : Vertex : ( 1, -3)

x - symmettry : x-1 =0

b) x=1

c) X intercept : y=0 :

0 = 2(x -1)^2 -3

(x-1)^2 = 3/2

x = 1 + /- sqrt(3/2)

d) y intercept : x=0

y = -1

e) Domain: all real ( -inf, inf)

range : ( -3, inf)