Answer:
a = -1/6
b = 7/6
c = 1
Step-by-step explanation:
[tex]f(x)=ax^3+2x^2+bx+c\\\\f(0)=a(0)^3+2(0)^2+b(0)+c\\\\1=c[/tex]
[tex]f(1)=a(1)^3+2(1)^2+b(1)+1\\\\4=a+2+b+1\\\\4=a+b+3\\\\1=a+b\\\\1-b=a[/tex]
[tex]f(-2)=(1-b)(-2)^3+2(-2)^2+b(-2)+1\\\\8=-8+8b+8-2b+1\\\\8=6b+1\\\\7=6b\\\\\frac{7}{6}=b[/tex]
[tex]1-b=a\\\\1-\frac{7}{6}=a\\ \\-\frac{1}{6}=a[/tex]
So, the true equation would be [tex]f(x)=-\frac{1}{6}x^3+2x^2+\frac{7}{6}x+1[/tex] if you substitute the coefficients and the constant.
