1. ## Domain +Graphing

I need to determine the domain, sketch the graph of each function and show/explain the connection between the graph and the domain:

g (x) = x^2
______________

x^2 + 2x - 3
x^2 + 2x-3 should be greater than 0 since (Any number)/0 is NOt Defined

(X+3)(X-1)>0
So either
X> -3 OR X>1

Thus domain becomes
(-3,1)u(1,Infinity)

Right?

Now how wld I graph and explain the connection??

x^2 + 2x-3 should be greater than 0 since (Any number)/0 is NOt Defined

(X+3)(X-1)>0
So either
X> -3 OR X>1

Thus domain becomes
(-3,1)u(1,Infinity)

Right?

Now how wld I graph and explain the connection??
The values of x such that $x^2+2x-3=0$ are forbidden. NOT where $x^2+2x-3$ is negative. There are three parts to the domain.

Wherever the denominator goes to zero will be a vertical asymptote (unless the numerator also goes to zero at this value of x, which does not happen here).

-Dan

3. Originally Posted by topsquark
The values of x such that $x^2+2x-3=0$ are forbidden. NOT where $x^2+2x-3$ is negative. There are three parts to the domain.

Wherever the denominator goes to zero will be a vertical asymptote (unless the numerator also goes to zero at this value of x, which does not happen here).

-Dan
Umm, So the domain becomes....??

And how would I be graphing this?

Umm, So the domain becomes....??
The domain is that "x" can be any number EXCEPT $-3,1$.

You can write it in three ways,
$x\in \mathbb{R}, x\not = -3,1$
Or,
$\left\{ \begin{array}{c}x<-3\\ -3
Or,
$x\in (-\infty,-3)\cup (-3,1) \cup (1,+\infty)$

Step 1 draw vertical line which are asymptotes for the undefinied values. That means draw $x=-3,x=1$