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

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: \[ y = x+1. \]