# Math Help - Difference Quotient Question

1. ## Difference Quotient Question

I could really use some help...a littel confused. I know that the difference quotient is f(x+h)-f(x)/h

The question in the assignment asked me to use the difference quotient to evaluate f(x)= x^2+13

this is how i did it, could you please let me know where I went wrong...

(x+h)(x+h)-x^2+13/h

x^2 +2hx+h^2-x^2 +13/h
the two x^2 are cancelled
the h's cancel, leaving 2x+h+13/2
my final anwer was 2x+h+13/h
I didn't get it correct and im not sure where I went wrong...
Thanks!

2. Originally Posted by epetrik
I could really use some help...a littel confused. I know that the difference quotient is f(x+h)-f(x)/h

The question in the assignment asked me to use the difference quotient to evaluate f(x)= x^2+13

this is how i did it, could you please let me know where I went wrong...

(x+h)(x+h)-x^2+13/h
1. f(x+h)= (x+h)^2+ 13, not (x+h)^2= (x+h)(x+h)
2. -f(x)= -(x^2+ 13)= -x^2- 13, not -x^2+ 13

x^2 +2hx+h^2-x^2 +13/h
the two x^2 are cancelled
the h's cancel, leaving 2x+h+13/2
my final anwer was 2x+h+13/h
I didn't get it correct and im not sure where I went wrong...
Thanks!

3. Originally Posted by epetrik
I could really use some help...a littel confused. I know that the difference quotient is f(x+h)-f(x)/h

The question in the assignment asked me to use the difference quotient to evaluate f(x)= x^2+13

this is how i did it, could you please let me know where I went wrong...

(x+h)(x+h)-x^2+13/h

x^2 +2hx+h^2-x^2 +13/h
the two x^2 are cancelled
the h's cancel, leaving 2x+h+13/2
my final anwer was 2x+h+13/h
I didn't get it correct and im not sure where I went wrong...
Thanks!
This is what I came up with:

$f(x)=x^2+13$

$f(x+h)=(x+h)^2$

$\frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h}=\frac{[(x-h)^2+13]-(x^2+13)}{h}=$

$\frac{x^2+2xh+h^2+13-x^2-13}{h}=\frac{h^2+2xh}{h}=\frac{h(h+2x)}{h}=\boxed{ h+2x}$