Question 1131288
{{{f(x)=x^(3)+4x^(2)-6x-13}}}

{{{a=-8}}}, {{{b=-4 }}}

{{{f(-8)=(-8)^(3)+4(-8)^(2)-6(-8)-13}}}
{{{f(-8)=-221}}}}

{{{f(-4)=(-4)^(3)+4(-4)^(2)-6(-4)-13}}}
{{{f(-4)=11}}}

Now we know:

at {{{x=-221}}}, the curve is {{{below}}} zero
at {{{x=11}}}, the curve is {{{above }}}zero

And, being a polynomial, the curve will be continuous, so somewhere in between the curve must cross through {{{f(x)=0}}}

Yes, there is a solution to {{{x^(3)+4x^(2)-6x-13= 0}}} in the interval [{{{-8}}}, {{{-4}}}]


{{{ graph( 600, 600, -15, 15, -10,10, x^(3)+4x^(2)-6x-13) }}}