Determine the oblique asymptote of the graph of the function. $$ g(x)=\frac{x^{2}+4 x-2}{x+3} $$

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1. We have the function 
   $$
     g(x)=\frac{x^{2}+4x-2}{x+3}.
   $$
   
2. To find the oblique asymptote, we perform polynomial long division of the numerator by the denominator.

3. Divide $$ x^{2} $$ (the leading term of the numerator) by $$ x $$ (the leading term of the denominator) to get $$ x $$. Multiply the divisor $$ x+3 $$ by $$ x $$ to obtain:
   $$
   x(x+3)=x^{2}+3x.
   $$

4. Subtract this product from the numerator:
   $$
   (x^{2}+4x-2)-(x^{2}+3x)=x-2.
   $$

5. Next, divide the new leading term $$ x $$ by $$ x $$ to obtain $$ 1 $$. Multiply the divisor $$ x+3 $$ by $$ 1 $$ to get:
   $$
   1\cdot(x+3)=x+3.
   $$

6. Subtract:
   $$
   (x-2)-(x+3)=-5.
   $$

7. The quotient from the division is $$ x+1 $$ and the remainder is $$-5$$. Thus, we can write:
   $$
   \frac{x^{2}+4x-2}{x+3}=x+1+\frac{-5}{x+3}.
   $$

8. As $$ x $$ tends to infinity, the term $$\frac{-5}{x+3}$$ approaches $$ 0 $$. Therefore, the graph of the function approaches the line:
   $$
   y=x+1.
   $$

Thus, the oblique asymptote is:
$$
y = x+1.
$$
The oblique asymptote of the function $$ g(x)=\frac{x^{2}+4x-2}{x+3} $$ is $$ y = x + 1 $$.

Determine the oblique asymptote of the graph of the function. \( g(x)=\frac{x^{2}+4 x-2}{x+3} \)

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