Question 949902
{{{f(x) = x^2 -x -6}}}...factor completely

{{{f(x) = x^2+2x -3x -6}}}

{{{f(x) = (x^2+2x) -(3x +6)}}}

{{{f(x) = x(x+2) -3(x +2)}}}

{{{f(x) = (x -3)(x + 2)}}}

{{{x}}} intercepts for this are:
{{{x -3 = 0}}}=> {{{x = 3}}}
and {{{x + 2 = 0}}}=>{{{x = -2}}}

To find the intercepts for each of the following  
a) {{{y=f(2x)}}}
b) {{{y=f(1/3x)}}}
c) {{{y=f(-3x)}}}

simply substitute the new {{{x}}} values

a)
{{{2x = 3}}} or {{{highlight(x = 3/2)}}} and {{{2x = -2}}} or {{{highlight(x = -1)}}}

b)
{{{(1/3)x = 3}}} or {{{highlight(x = 9)}}} and {{{(1/3)x = -2}}} or {{{highlight(x =-6)}}}

c)
{{{-3x = 3}}} or {{{highlight(x = -1)}}} and {{{-3x = -2}}} or {{{highlight(x = 2/3)}}}


you can also do it this way:

a)

{{{y=f(2x)=(2x)^2-2x-6}}}
{{{y= 4x^2-2x-6}}}
set {{{y}}} to zero to find the x-intercepts
{{{0= 4x^2-2x-6}}}...........both sides divide by {{{2}}}
{{{0= 2x^2-x-3}}}...........factor
{{{2x^2+2x-3x-3=0}}}
{{{(2x^2+2x)-(3x+3)=0}}}
{{{2x(x+1)-3(x+1)=0}}}
{{{(x+1) (2x-3)=0}}}
if {{{(x+1) =0 }}}=> {{{x=-1}}}
if {{{(2x-3)=0}}}=>{{{x=3/2}}}
the x-intercepts are {{{x=-1}}} and {{{x=3/2 }}}

same way you can do b) and c)