Determine the oblique asymptote of the graph of the function.

Ask by Graham Chang. in the United States

Mar 20,2025

Question

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

UpStudy AI · Accepted Answer

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.

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