Graph all asymptotes of the rational function. $$ f(x)=\frac{x^{2}-5 x+1}{x-3} $$

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**Step 1. Find the Vertical Asymptote**

The function is
$$
f(x)=\frac{x^2-5x+1}{x-3}.
$$
Set the denominator equal to zero:
$$
x-3=0 \quad \Rightarrow \quad x=3.
$$
Since the numerator $$x^2-5x+1$$ is not zero at $$x=3$$ (check: $$3^2-5\cdot3+1=9-15+1=-5$$), there is a vertical asymptote at
$$
x=3.
$$

**Step 2. Find the Oblique Asymptote**

Since the degree of the numerator ($$2$$) is one more than the degree of the denominator ($$1$$), the function has an oblique (or slant) asymptote. We perform polynomial division of $$x^2-5x+1$$ by $$x-3$$.

1. **Divide $$x^2$$ by $$x$$:**  
   $$
   \frac{x^2}{x}=x.
   $$
   Multiply the divisor by $$x$$:
   $$
   x(x-3)=x^2-3x.
   $$
   Subtract from the original numerator:
   $$
   (x^2-5x+1) - (x^2-3x) = -2x+1.
   $$

2. **Divide $$-2x$$ by $$x$$:**  
   $$
   \frac{-2x}{x}=-2.
   $$
   Multiply the divisor by $$-2$$:
   $$
   -2(x-3)=-2x+6.
   $$
   Subtract:
   $$
   (-2x+1)-(-2x+6)=1-6=-5.
   $$

Thus, the result of the division is:
$$
f(x)=x-2+\frac{-5}{x-3}.
$$
As $$x$$ approaches infinity (or negative infinity), the term $$\frac{-5}{x-3}$$ approaches $$0$$. Therefore, the oblique asymptote is
$$
y=x-2.
$$

**Answer**

The rational function has the following asymptotes:
- A vertical asymptote at $$x=3$$.
- An oblique asymptote at $$y=x-2$$.
The function has a vertical asymptote at $$x=3$$ and an oblique asymptote at $$y=x-2$$.

Graph all asymptotes of the rational function.

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