# Thread: Stuck with a Binomial using the Difference Quotient

1. ## Stuck with a Binomial using the Difference Quotient

The problem is:

$f(x)= -5+5x^2$

I already know that it will end as 10x (it's derivative by the way).

But using the difference quotient:

$\frac{f(x+h)-f(x)}{h}$

I'm told that,

$f(x)= -5+5x^2$

Can be written in the form

$Ah+Bx+C$

Using the Calculus short cuts. I find that "B" is (10) and I find that "C" is (0)

However I don't know what what "A" is.

And when I do it long hand, I get stuck at this point.

$\frac{5(2x^2+2xh+h^2)}{h}$

How do I get "10x" from that above and find "A"?

2. Originally Posted by Zanderist
The problem is:

$f(x)= -5+5x^2$

I already know that it will end as 10x (it's derivative by the way).

But using the difference quotient:

$\frac{f(x+h)-f(x)}{h}$

I'm told that,

$f(x)= -5+5x^2$

Can be written in the form

$Ah+Bx+C$

Using the Calculus short cuts. I find that "B" is (10) and I find that "C" is (0)

However I don't know what what "A" is.

And when I do it long hand, I get stuck at this point.

$\frac{5(2x^2+2xh+h^2)}{h}$

How do I get "10x" from that above and find "A"?
I am supposing that the full question is asking you to find the derivative with the difference quotient?

Something to keep in mind is that the secondary form $Ah+Bx+C$ may come from plugging into the quotient. I prefer to use the function itself.

Considering your function is $f(x) = -5 + 5x^2$

Plug in for $f(x+h) = -5 + 5(x+h)^2$; This goes into your difference quotient. Expand and simplify.

3. I figured out what I had done wrong.

I didn't distribute the negative sign fully.

"A" for this problem is (5).