# Thread: Find the difference quotient and simplify

1. ## Find the difference quotient and simplify

Find the difference quotient and simplify answer: f(5+h)-f(5)/h, h cannot = 0
when f(p)=6p^2-4p.

ok this is what i did and i got stuck.

f(5+h)= 6(5+h)^2-4(5+h)

=6(25+10h+h^2)-20-4h

=150+60h+6h^2-20-4h

= 130+56h+6h^2

??????? help me out...its getting so late can't keep my eyes open lol

2. Originally Posted by refresh
Find the difference quotient and simplify answer: f(5+h)-f(5)/h, h cannot = 0
when f(p)=6p^2-4p.

ok this is what i did and i got stuck.

f(5+h)= 6(5+h)^2-4(5+h)

=6(25+10h+h^2)-20-4h

=150+60h+6h^2-20-4h

= 130+56h+6h^2

??????? help me out...its getting so late can't keep my eyes open lol
You're right so far. Now find $f(5)$, substitute both $f(5), \ f(5+h)$ into the formula and simplify.

3. The function $f$ is defined as follows :

$f(p) = 6p^2 - 4p$

Basically, you are trying to simplify this :

$\frac{f(5+h)-f(5)}{h}$ , $h \neq 0$

It is equivalent to (by substituting the function into your expression) :

$\frac{[6(5 + h)^2 - 4(5 + h)] - [6(5)^2 - 4(5)]}{h}$ , $h \neq 0$

Expand this :

$\frac{[6(5^2 + 10h + h^2) - 20 - 4h] - (150 - 20)}{h}$ , $h \neq 0$

Further expansion required ...

$\frac{150 + 60h + 6h^2 - 20 - 4h - 130}{h}$ , $h \neq 0$

Simplify :
> get rid of the constants (150, 130, 20, ...), they cancel out (who would have guessed it ?).
> simplify the linear coefficients (60h - 4h) !
> reorganize the equation, looks better ...

$\frac{6h^2 + 56h}{h}$ , $h \neq 0$

Factorize with $h$ :

$\frac{h(6h + 56)}{h}$ , $h \neq 0$

Finally, cancel out the $h$, to find :

$6h + 56$

And you are done

EDIT : ah Defunkt you win, I was busy checking my reply :/

4. Originally Posted by Bacterius
The function $f$ is defined as follows :

$f(p) = 6p^2 - 4p$

Basically, you are trying to simplify this :

$\frac{f(5+h)-f(5)}{h}$ , $h \neq 0$

It is equivalent to (by substituting the function into your expression) :

$\frac{[6(5 + h)^2 - 4(5 + h)] - [6(5)^2 - 4(5)]}{h}$ , $h \neq 0$

Expand this :

$\frac{[6(5^2 + 10h + h^2) - 20 - 4h] - (150 - 20)}{h}$ , $h \neq 0$

Further expansion required ...

$\frac{150 + 60h + 6h^2 - 20 - 4h - 130}{h}$ , $h \neq 0$

Simplify :
> get rid of the constants (150, 130, 20, ...), they cancel out (who would have guessed it ?).
> simplify the linear coefficients (60h - 4h) !
> reorganize the equation, looks better ...

$\frac{6h^2 + 56h}{h}$ , $h \neq 0$

Factorize with $h$ :

$\frac{h(6h + 56)}{h}$ , $h \neq 0$

Finally, cancel out the $h$, to find :

$6h + 56$

And you are done

EDIT : ah Defunkt you win, I was busy checking my reply :/
omg i thought i had to substitute the f(5) into 130+56h+6h^2 lol no wonder i been getting the wrong answer. thank you so much.