Could any one please explain to me how to solve them thanks =)
I'll help with Q2: $\displaystyle y = \frac{x^2 - 4}{2x^2 + 5x - 12}$.
x -int: Apply the definition: Solve $\displaystyle x^2 - 4 = 0$.
Vertical asymptote: Apply the definition: Solve $\displaystyle 2x^2 + 5x - 12 = 0$.
Horizontal asymptote: Apply the definition and consider the limit as $\displaystyle x \rightarrow \pm \infty$.
Two options:
1. From polynomial long division $\displaystyle y = \frac{x^2 - 4}{2x^2 + 5x - 12} = \frac{1}{2} - \frac{5x-4}{2(2x^2 + 5x - 12)}$.
2. $\displaystyle y = \frac{x^2 - 4}{2x^2 + 5x - 12} = \frac{1 - \frac{4}{x^2}}{2 + \frac{5}{x} - \frac{12}{x^2}}$.
Either way, $\displaystyle y \rightarrow 2$ ....