algebra 2 help

**Steven2** · January 4th 2009, 05:27 PM

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

**mr fantastic** · January 4th 2009, 05:50 PM

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, $y \rightarrow 2$ ....

**Steven2** · January 4th 2009, 06:13 PM

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.

**mr fantastic** · January 4th 2009, 06:24 PM

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.

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