Question 964657
the zeros of the function {{{ (x-3)(x+3)(x-4) =0}}} are:

{{{ (x-3)=0}}}=>{{{ x=3}}}
{{{ (x+3)=0}}}=>{{{ x=-3}}}
{{{ (x-4)=0}}}=>{{{ x=4}}}

function is:

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

{{{f(x)= (x^2-9)(x-4) }}}

{{{f(x)= x^3-9x-4x^2+36 }}}

{{{f(x)= x^3-4x^2-9x+36 }}}

you have three points already, find some more, plot them and draw a graph

{{{x}}}|{{{f(x)=y}}}
{{{4}}}|{{{0}}}
{{{-3}}}|{{{0}}}
{{{3}}}|{{{0}}}
{{{1}}}|{{{24}}}...{{{f(1)= 1^3-4*1^2-9*1+36= 24}}}
{{{2}}}|{{{10}}}...{{{f(2)= 2^3-4*2^2-9*2+36= 10}}}
{{{3}}}|{{{0}}}...{{{f(3)= 3^3-4*3^2-9*3+36= 0}}}


{{{drawing( 600, 600, -10, 10, -10, 30, 
circle(4,0,.12),circle(-3,0,.12),circle(3,0,.12),
circle(1,24,.12),circle(2,10,.12),circle(3,0,.12),
 graph( 600, 600, -10, 10, -10, 30, x^3-4x^2-9x+36 )) }}}