Find a polynomial of the formf(x) = ax3 + bx2 + cx + dsuch that f(0) = −3, f(1) = 2, f(3) = 5, and f(4) = 0. (A graphing calculator is recommended.)

Answer:[tex]f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3[/tex]Step-by-step explanation:We are given that [tex]f(x)=ax^3+bx^2+cx+d[/tex]f(0)=-3,f(1)=2,f(3)=5 and f(4)=0We have to find the polynomialSubstitute the value x=0 then ,we get f(0)=d=-3Substitute x=1 then we get [tex]a+b+c-3=2[/tex][tex]a+b+c=2+3=5[/tex][tex]a+b+c=5[/tex] (equation I)Substitute x=3 then we get [tex]a(3)^3+b(3)^2+c(3)-3=5[/tex][tex]27a+9b+3c=5+3=8[/tex][tex]27a+9b+3c=8[/tex]  (Equation II)Substitute x=4 then we get[tex]a(4)^3+b(4)^2+c(4)-3=0[/tex][tex]64a+16b+4c=3[/tex] (Equation III)Equation I multiply by 3 then subtract from equation II[tex]24a+6b=-7[/tex] (Equation IV)Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III[tex]84a+12b=-23[/tex] (Equation V)Equation IV multiply by 2 and then subtract from equation V[tex]36a=-9[/tex][tex]a=-\frac{9}{36}[/tex][tex]a=-\frac{1}{4}[/tex]Substitute the value of a in equation IV then we get [tex]24(-\frac{1}{4})+6b=-7[/tex][tex]-6+6b=-7[/tex][tex]6b=-7+6=-1[/tex][tex]b=-\frac{1}{6}[/tex]Substitute the value of b in equation I then we get [tex]-\frac{1}{4}-\frac{1}{6}+ c=5[/tex][tex]-\frac{5}{12}+c=5[/tex][tex]c=5+\frac{5}{12}=\frac{65}{12}[/tex]Substitute the values then we get [tex]f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3[/tex]