



 
Hi, I am going to change the function slightly as I think you have copied it incorrectly. I want the expression to be \[f(x) = \left\{ \begin{array}{ll}\frac{x^2 + 3x + 2}{x + 2} & \mbox{if $x \neq 2$}\\4 & \mbox{if $x = 2$}. \end{array} \right. \] The value of $2$ is important since the expression \[\frac{x^2 + 3x + 2}{x + 2}\] doesn't exist at $x = 2$ since the denominator is zero. Looking at the expression \[\frac{x^2 + 3x + 2}{x + 2}\] more closely you can see that the numerator factors so that \[\frac{x^2 + 3x + 2}{x + 2} = \frac{(x + 2)(x + 1)}{x + 2}.\] Thus using cancellation it seems that the expressions $\large \frac{x^2 + 3x + 2}{x + 2}$ and $x + 1$ are equal but that is not true since the first expression doesn't exist for $x = 2$ and the second expression has value $1$ for $x = 2.$ Now back to the problem of the domain and graph of \[f(x) = \left( \begin{array}{ll}\frac{x^2 + 3x + 2}{x + 2} & \mbox{if $x \neq 2$}\\4 & \mbox{if $x = 2$}. \end{array} \right. \] If $x \neq 2$ then $f(x)$ is evaluated using $\large \frac{x^2 + 3x + 2}{x + 2}$ and for $x = 2, \; f(x) = f(2) = 4$ and thus $f(x)$ exists for all real numbers $x$ and thus the domain of $f(x)$ is all real numbers. To graph $y = f(x)$ I would start by graphing $y = x + 1$ which is the line with slope 1 and passing through $(0, 1).$ y = x + 1 We know that the functions $y = x + 1$ and $y = f(x)$ have the same values for all $x$ except $x = 2.$ At $x = 2$ we have $f(x) = f(2) = 4$ and thus my graph of $y = f(x) is y = f(x) The small circle on the line at $x = 2$ is to indicate that there is a "hole" in the line there and the value of $f(2)$ is 4. I hope this helps,  


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