Question 930999
A cubic polynomial function:

{{{ f(x) =ax^3+bx^2+cx+d}}}

given:
has leading coefficient {{{ -2 }}}=>{{{highlight(a=-2)}}}

so far we know {{{ f(x) =-2x^3+bx^2+cx+d}}}

and intercepts the y-axis at {{{2}}}: => {{{highlight(d=2)}}}=> {{{ f(x) =-2x^3+bx^2+cx+2}}}

If {{{f(1)=1}}} =>  {{{ f(1) =-2(1)^3+b(1)^2+c(1)+2=1}}}

=>{{{ -2+b+c+2=1}}}

=>{{{ b+c=1}}}.......eq.1


 and {{{f(-2) =-2}}}  =>  {{{ f(-2) =-2(-2)^3+b(-2)^2+c(-2)+2=-2}}}

=>  {{{ -2(-8)+b(4)+c(-2)+2=-2}}}

=>{{{16+4b-2c+2=-2}}}

=>{{{ 4b-2c=-20}}}.......eq.2

solve the system:

{{{ b+c=1}}}.......eq.1

{{{ 4b-2c=-20}}}.......eq.2
_________________________

{{{ b+c=1}}}.......eq.1 ...solve for {{{b}}}

{{{ b=1-c}}}...substitute in eq.2

{{{ 4(1-c)-2c=-20}}}.......eq.2...solve for {{{c}}}

{{{ 4-4c-2c=-20}}}

{{{ 4-6c=-20}}}

{{{ 4+20=6c}}}

{{{ 24=6c}}}

{{{ 24/6=c}}}

{{{highlight(c=4)}}}

find {{{b}}}

{{{ b=1-c}}}
{{{ b=1-4}}}
{{{highlight( b=-3)}}}

 the complete function is:

{{{ f(x) =-2x^3-3x^2+4x+2}}}


find {{{f(-1)}}} :

{{{ f(-1) =-2(-1)^3-3(-1)^2+4(-1)+2}}}

{{{ f(-1) =-2(-1)-3(1)+4(-1)+2}}}

{{{ f(-1) =2-3-4+2}}}

{{{ f(-1) =-3}}}


check:
{{{ f(-2) =-2(-2)^3-3(-2)^2+4(-2)+2}}}

{{{ f(-2) =-2(-8)-3(4)+4(-2)+2}}}

{{{ f(-2) =16-12-8+2}}}

{{{ f(-2) =4-8+2}}}

{{{ f(-2) =6-8}}}

{{{ f(-2) =-2}}}...as given


{{{drawing( 600, 600, -10,10, -10, 10,circle(1,1,.15),locate(1,1,p(1,1)),circle(0,2,.15),locate(0,2,p(0,2)),circle(-2,-2,.15),locate(-2,-2,p(-2,-2)), graph( 600, 600, -10,10, -10, 10, -2x^3-3x^2+4x+2)) }}}