Thanks Heuraka, I am going to take a different path.
y= 3x+3
The y intercept is where x=0 so 3x=0 put your finger over 3x you don't need that bit!
y=3
-----------------
The x intercept is where y=0 so you have to solve
0=3x+3
-3=3x
x=-1
-----------------------
Now you have your 2 intercepts it is really easy to graph.
find the intercepts and use them to graph the equation
$$\begin{array}{rcl}
y & = & 3x+3 \quad | \quad -3x\\ \\
y-3x &=& 3 \quad | \quad :3\\ \\
\frac{y-3x}{3} &=&1\\\\
\frac{y}{3} + \frac{-3x}{3} &=&1\\\\
\frac{y}{3} + \frac{-x}{1} &=&1\\\\
\frac{y}{3} + \frac{x}{-1} &=&1\\\\
\frac{x}{-1} + \frac{y}{3} &=&1
\end{array}$$
$$\boxed{\dfrac{x}{ x_{intercept} } + \dfrac{y}{ y_{intercept} } &=&1 }\\\\
x_{intercept} = -1\\
y_{intercept} = 3$$
Thanks Heuraka, I am going to take a different path.
y= 3x+3
The y intercept is where x=0 so 3x=0 put your finger over 3x you don't need that bit!
y=3
-----------------
The x intercept is where y=0 so you have to solve
0=3x+3
-3=3x
x=-1
-----------------------
Now you have your 2 intercepts it is really easy to graph.