Complete the square to put the quadratic in standard form...

Question

f(x) = x² + 2x -8

Vertex:

Concave:

Y/X intercept:

Domain/range:

Increasing/decreasing intervals:

Determine intervals where f(x) > 0 and where f(x) < 0

Answer 1

x^2 +2x -8

= x^2 +2x + 1 -8-1

= (x+1)^2 -9

vertex is (-1,-9)

It's an upward opening parabola which is concave up

slope is everywhere increasing

y intercept = -8 or the point (0,-8)

x^2+2x-8 = (x+4)(x-2)

x intercepts are x=2 and -4 or the points (2,0) and (-4,0)

decreasing interval is -infinity to -1

increasing interval is -1 to +infinity

f(x)<0 when -4<x<2

f(x)>0 everywhere else, when -infinity<x<-4 and when 2<x<+infinity