Question 15182
okay {{{f(x)=6x^2 + 5x - 4 }}}

x is just the dependent variable. you can just as easily say 

{{{f(a)=6a^2 + 5a - 4}}}

or 

{{{f(2)=6(2)^2 + 5(2) - 4}}} 

That's just input. Now, the problem you are asking is VERY important for later mathematics, so spend some time trying to understand it for sure. 

{{{ (f(x+h)-f(x)) / h}}} 

so 
{{{ ((6(x+h)^2 + 5(x+h) - 4)-(6x^2 + 5x - 4))/h }}}

{{{ ( 6(x^2 + 2xh + h^2) + 5x + 5h - 4 - 6x^2 - 5x + 4)/h }}}

REMEBER TO DISTRIBUTE NEGATIVE IN SECOND TERM!
{{{ ( 6x^2+12xh+6h^2 + 5x + 5h - 4 - 6x^2 - 5x + 4)/h }}}
Combined any common terms (6x^2-6x^2=0, 5x-5x=0 -4+4=0)
{{{ ( 12xh + 6h^2 + 5h)/h }}}
Factor out and cancel out h
{{{ ( h(12x + 6h + 5))/h }}}

{{{  12x + 6h + 5 }}}

Like I said, this is a very important thing for later mathematical concept. It's called the difference quotient.