+0  
 
0
44
1
avatar

Y = 2x^2 - 4x - 6 

 

a.) convert to vertex form 

b.) find x- intercept. Solutions must be exact

c.) find all values such that y > 0 

d.) find the range of dunction

e.) when is the function decreasing? 

 Dec 3, 2018
 #1
avatar+15350 
+1

1

     2 (x^2-2x) -6   complete the square

Y=  2 (x-1)^2 - 8

 

2

  x intercept(s) happen when y = 0   set the equation to y=0

  0 = 2(x-1)^2 - 8

4 = (x-1)^2

+- 2 = x-1

x = 3, -1

 

3

  From 2 above    (-inf,-1) U (3, +inf)

 

Range = all values of x are allowed in the domain...the range extends from the vertex at -8 to + inf     [-8, +inf)

This is an upright parabola....the funtion is decreasing from -inf to the vertex (1,-8)    or for values of x  ( -inf ,1]

 Dec 3, 2018
edited by ElectricPavlov  Dec 3, 2018

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