Question 1134798


{{{y=h(x)}}}, where {{{y=x^2+x}}}. 

The graph has been transformed. The new equation is {{{(1/2)y=h(2x)}}}. 

{{{(1/2)(x^2+x)=h(2x)}}}......since {{{y=h(x)}}}=>{{{h(x)=x^2+x}}}

then {{{h(2x)=(2x)^2+2x}}}

{{{h(2x)=4x^2+2x}}}

and we have

{{{(1/2)(x^2+x)=4x^2+2x}}}

{{{x^2+x=2*(4x^2+2x)}}}

{{{x^2+x=highlight(8x^2+4x)}}}


answer:

D. {{{8x^2+4x}}}