Question 1139683


 Suppose that the x intercepts of the graph of {{{y=(x)}}} are {{{-8}}} and {{{1}}}. What are the x intercepts of the graph of {{{y=(x+4)}}}?


notice that the {{{y }}}coordinate is always {{{zero}}} for the {{{x }}}intercept and the {{{x }}}coordinate is always {{{zero}}} for the {{{y }}}intercept. 

That is, the {{{x }}} intercept is the point ({{{x}}},{{{0}}}).

if the graph of {{{y=(x)}}} are {{{-8}}} and {{{1}}},  the points are ({{{-8}}},{{{0}}}) and ({{{1}}},{{{0}}})


and, the {{{x }}}intercepts of the graph of{{{ y=(x+4)}}} will be where {{{(x+4)=-8}}} and {{{(x+4)=1}}}

{{{x+4=-8}}}
{{{x=-8-4}}}
{{{x=-12}}}

and

{{{x+4=1}}}
{{{x=1-4}}}
{{{x=-3}}}

so, the {{{x }}}intercepts of the graph of {{{y=(x+4)}}} are{{{ -12}}} and {{{-3}}}
 
or at the points  ({{{-12}}},{{{0}}}) and ({{{-3}}},{{{0}}})


or do it this way:, using zero product theorem

factors of the function {{{f ( x ) }}}are:
 
{{{f ( x ) = ( x + 8 ) ( x - 1 )}}}...x - intercepts of this function are {{{- 8}}} and {{{1}}}

multiply to expand

{{{f ( x )  = x^2 + 7 x - 8}}}

now find {{{f ( x + 4 ) }}}

{{{f ( x + 4 ) =(x+4)^2 + 7 (x+4) - 8}}}

={{{x^2 + 8 x+16+7x+28 - 8}}}

={{{x^2 +15x+36}}}..........factor completely

={{{x^2 +3x+12x+36}}}

={{{(x^2 +12x)+(3x+36)}}}

={{{x(x +12)+3(x+12)}}}

 ={{{(x + 12) (x + 3)}}}

so, the {{{x}}} - intercepts of the graph {{{f ( x + 2 )}}} are {{{- 12}}} and {{{-3}}}