To find the slant asymptote of the given function:

$f(x) = \frac{x(x+8)}{\sqrt{x} + 2},$

we proceed as follows:

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Step 1: Check the degree of numerator and denominator

The numerator $x(x + 8) = x^2 + 8x$ has a degree of 2 (the highest power of $x$), and the denominator $\sqrt{x} + 2$ behaves like $\sqrt{x}$ for large $x$. The numerator grows faster than the denominator for large $x$, indicating the presence of a slant asymptote.

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Step 2: Perform long division

We divide $x^2 + 8x$ by $\sqrt{x} + 2$.

Division process:

1. Rewrite the numerator: $x^2 + 8x$.
2. Divide the first term: $x^2 \div \sqrt{x} = x^{3/2}$.
3. Multiply $x^{3/2}$ by $\sqrt{x} + 2$: $x^{3/2}(\sqrt{x} + 2) = x^2 + 2x^{3/2}$.
4. Subtract: $(x^2 + 8x) - (x^2 + 2x^{3/2}) = -2x^{3/2} + 8x.$
5. Divide the leading term: $-2x^{3/2} \div \sqrt{x} = -2x$.
6. Multiply $-2x$ by $\sqrt{x} + 2$: $-2x(\sqrt{x} + 2) = -2x^{3/2} - 4x$.
7. Subtract: $(-2x^{3/2} + 8x) - (-2x^{3/2} - 4x) = 12x.$

The quotient so far is $x^{3/2} - 2x$, and the remainder term ( \frac{