1. ## algebra 2 help

Could any one please explain to me how to solve them thanks =)

2. Originally Posted by Steven2
Could any one please explain to me how to solve them thanks =)

I'll help with Q2: $y = \frac{x^2 - 4}{2x^2 + 5x - 12}$.

x -int: Apply the definition: Solve $x^2 - 4 = 0$.

Vertical asymptote: Apply the definition: Solve $2x^2 + 5x - 12 = 0$.

Horizontal asymptote: Apply the definition and consider the limit as $x \rightarrow \pm \infty$.

Two options:

1. From polynomial long division $y = \frac{x^2 - 4}{2x^2 + 5x - 12} = \frac{1}{2} - \frac{5x-4}{2(2x^2 + 5x - 12)}$.

2. $y = \frac{x^2 - 4}{2x^2 + 5x - 12} = \frac{1 - \frac{4}{x^2}}{2 + \frac{5}{x} - \frac{12}{x^2}}$.

Either way, $y \rightarrow 2$ ....

3. Thanks for the response but could you help me factor the denominator.

Example X^2-9x+18 factors out to (x-3)(X-6)

Just cant seem to factor this problem.

4. Originally Posted by Steven2
Thanks for the response but could you help me factor the denominator.

Example X^2-9x+18 factors out to (x-3)(X-6)

Just cant seem to factor this problem.
$2x^2 + 5x - 12 = (2x-3)(x+4)$.

If you had trouble factorising then you should have tried solving the equation using the quadratic formula.