Question 916789


The old graph was {{{y = f(x)}}}. 

The new graph, after transformation, is {{{y}}}' = {{{3f}}}({{{x}}}'- {{{2}}})+ {{{4 }}}, which can also be written as ({{{y}}}' -{{{4}}})/{{{3}}} = {{{f}}}({{{x}}}' - {{{2}}}). 

(Here the accent sign ' denotes "new coordinate"; and not differentation or derivative of a variable or anything.) 

Hence the transformation is 

{{{x}}} = {{{x}}}' -{{{2 }}}
{{{y}}} = ({{{y}}}' -{{{ 4}}})/{{{3}}} 

The old coordinates were ({{{x}}},{{{ y}}}) = ({{{4}}},{{{-6}}}). 

So the new coordinates ({{{x}}}', {{{y}}}') are found by solving 

{{{4}}}= {{{x}}}' - {{{2}}}  => {{{4+2}}}= {{{x}}}' => {{{6}}}= {{{x}}}' 

{{{-6}}}= ({{{y}}}' -{{{4}}})/{{{3}}} =>{{{-6*3}}}={{{y}}}' -{{{4}}}=>{{{-18}}}={{{y}}}' -{{{4}}} =>{{{-18+4}}}={{{y}}}' =>{{{-14}}}={{{y}}}' 

Hence the new coordinates are ({{{x}}}',{{{ y}}}') = ({{{6}}}, {{{-14}}}).