Find all vertical asymptotes of the following function. $$ f(x)=\frac{2 x^{2}-17 x+8}{2 x^{2}-18 x} $$

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To find the vertical asymptotes of the function

$$
f(x) = \frac{2x^2 - 17x + 8}{2x^2 - 18x},
$$

we need to determine where the denominator is equal to zero, as vertical asymptotes occur at these points, provided the numerator is not also zero at those points.

First, we set the denominator equal to zero:

$$
2x^2 - 18x = 0.
$$

We can factor out a common term:

$$
2x(x - 9) = 0.
$$

Setting each factor equal to zero gives us:

1. $$ 2x = 0 $$ which simplifies to $$ x = 0 $$,
2. $$ x - 9 = 0 $$ which simplifies to $$ x = 9 $$.

Thus, the potential vertical asymptotes are at $$ x = 0 $$ and $$ x = 9 $$.

Next, we need to check if the numerator is also zero at these points to confirm that they are indeed vertical asymptotes.

1. For $$ x = 0 $$:

$$
2(0)^2 - 17(0) + 8 = 8 \neq 0.
$$

Since the numerator is not zero, there is a vertical asymptote at $$ x = 0 $$.

2. For $$ x = 9 $$:

$$
2(9)^2 - 17(9) + 8 = 2(81) - 153 + 8 = 162 - 153 + 8 = 17 \neq 0.
$$

Since the numerator is also not zero, there is a vertical asymptote at $$ x = 9 $$.

In conclusion, the vertical asymptotes of the function $$ f(x) $$ are:

$$
\boxed{0 \text{ and } 9}.
$$
The vertical asymptotes are at $$ x = 0 $$ and $$ x = 9 $$.

Find all vertical asymptotes of the following function.

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