Question 1188473



given:

 zeros at {{{-6}}}, {{{0}}}, and {{{4}}}
{{{f(-7)=-8}}}


{{{f(x) =a(x-x[1])(x-x[2])(x-x[3])}}}

{{{f(x) =a(x-(-6))(x-0)(x-4)}}}

{{{f(x) =a(x+6)(x)(x-4)}}}

{{{f(x) =a(x^3 + 2x^2 - 24x)}}}...........substitute {{{f(-7)=-8}}}

{{{-8=a((-7)^3 + 2(-7)^2 - 24(-7))}}}

{{{-8=a(-77)}}}

{{{a=-8/(-77)}}}

{{{a=8/77}}}

{{{f(x) =(8/77)(x^3 + 2x^2 - 24x)}}}

{{{f(x) =(8/77) x^3+ (16/77)x^2 - (192/77)x}}}


{{{ graph( 600, 600, -10, 10, -10, 10, (8/77) x^3+ (16/77)x^2 - (192/77)x) }}}