Tập xác định: \(D = \mathbb{R}\backslash \left\{ { - 5} \right\}\)
Ta có: \(a = \mathop {\lim }\limits_{x \to + \infty } \frac{y}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{2{x^2} - 3x}}{{x + 5}}}}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2{x^2} - 3x}}{{{x^2} + 5x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 - \frac{3}{x}}}{{1 + \frac{5}{x}}} = 2\)
\(b = \mathop {\lim }\limits_{x \to + \infty } (y - ax) = \mathop {\lim }\limits_{x \to + \infty } (\frac{{2{x^2} - 3x}}{{x + 5}} - 2x) = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 13x}}{{x + 5}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 13}}{{1 + \frac{5}{x}}} = - 13\)
Ta có: \(\mathop {\lim }\limits_{x \to + \infty } [y - (ax + b)] = \mathop {\lim }\limits_{x \to + \infty } [y - (2x - 3)] = \mathop {\lim }\limits_{x \to + \infty } \frac{{2{x^2} - 3x}}{{x + 5}} - (2x - 13) = \mathop {\lim }\limits_{x \to + \infty } \frac{{65}}{{x + 5}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{65}}{x}}}{{1 + \frac{5}{x}}} = 0\)
Do đó, đồ thị hàm số có tiệm cận xiên là đường thẳng y = 2x - 13