Question 1206114
{{{f(x)= ax^3+bx^2+cx+d}}}

if
{{{f(0)=-5}}} => {{{x=0}}} and {{{f(x)=-5}}}

{{{-5= a*0^3+b*0^2+c*0+d}}}
{{{d=-5}}}.....eq.1

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


{{{f(-1)=-6}}} => {{{x=-1}}} and {{{f(x)=-6}}}

{{{-6= a*(-1)^3+b*(-1)^2+c*(-1)-5}}}

{{{-6+5= -a+b-c}}}....solve for{{{ a}}}

{{{a=b-c+1}}}.....eq.2


{{{f(3)=4}}}=> {{{x=3}}} and {{{f(x)=4}}}

{{{4= a*(3)^3+b*(3)^2+c*(3)-5}}}

{{{4+5=27a+9b+3c}}}....solve for {{{a}}}

{{{27a=9-9b-3c}}}

{{{a=9/27-9b/27-3c/27}}}

{{{a=1/3-b/3-c/9}}}.....eq.3



 {{{f(4)=-4}}}=> {{{x=4}}} and {{{f(x)=-4}}}

{{{-4= a*(4)^3+b*(4)^2+c*(4)-5}}}

{{{-4+5=64a  + 16b + 4c }}}

{{{a = 1/64 (-16b - 4c + 1)}}}

{{{a=-b/4 - c/16 + 1/64}}}.......eq.4


from eq.2 and eq.3 we have

{{{b-c+1=1/3-b/3-c/9}}}....solve for {{{c}}}

{{{c = (3 b)/2 + 3/4}}}.....eq.5


from eq.2 and eq.4 we have

{{{b-c+1=-b/4 - c/16 + 1/64}}}

{{{c = (4 b)/3 + 21/20}}}.....eq.6


from eq.5 and 6 we have

{{{(3 b)/2 + 3/4=(4 b)/3 + 21/20}}}...solve for {{{b}}}

{{{b = 9/5}}}


go to eq.5, substitute {{{b}}}

{{{c = (3 (9/5))/2 + 3/4}}}.....eq.5

{{{c=69/20}}}


go to eq.2, substitute {{{b}}}, and {{{c}}}

{{{a=9/5-69/20+1}}}.....eq.2

{{{a=-13/20}}}


your function is

{{{f(x)= (-13/20)x^3+(9/5)x^2+(69/20)x-5}}}