$f(x) = \frac{x^2+2x-15}{x^2-7x-18}$ is negative if and only if

When is $f(x) = \frac{x^2+2x-15}{x^2-7x-18}$ negative?

To find when a rational function is negative, we need to determine when the fraction $\frac{\text{numerator}}{\text{denominator}} < 0$.

A fraction is negative when the numerator and denominator have opposite signs.

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Factor the numerator and denominator

Factoring the numerator: $x^2 + 2x - 15$

We need two numbers that multiply to $-15$ and add to $2$.

These numbers are $5$ and $-3$ (since $5 \times (-3) = -15$ and $5 + (-3) = 2$)

Therefore: $x^2 + 2x - 15 = (x + 5)(x - 3)$

Factoring the denominator: $x^2 - 7x - 18$

We need two numbers that multiply to $-18$ and add to $-7$.

These numbers are $-9$ and $2$ (since $(-9) \times 2 = -18$ and $(-9) + 2 = -7$)

Therefore: $x^2 - 7x - 18 = (x - 9)(x + 2)$

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Rewrite the function

$f(x) = \frac{(x + 5)(x - 3)}{(x - 9)(x + 2)}$

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Find critical points

Critical points are values where the function equals zero or is undefined:

Zeros (numerator = 0): $x = -5$ and $x = 3$

Undefined points (denominator = 0): $x = -2$ and $x = 9$

These points divide the number line into intervals: $(-\infty, -5)$, $(-5, -2)$, $(-2, 3)$, $(3, 9)$, and $(9, \infty)$

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Test the sign in each interval

For each interval, we check the sign of each factor:

Interval	$(x+5)$	$(x-3)$	$(x-9)$	$(x+2)$	Overall Sign
$(-\infty, -5)$	$-$	$-$	$-$	$-$	$\frac{(-)(-)]}{(-)(-)} = +$
$(-5, -2)$	$+$	$-$	$-$	$-$	$\frac{(+)(-)}{(-)(-)} = -$
$(-2, 3)$	$+$	$-$	$-$	$+$	$\frac{(+)(-)}{(-)(+)} = +$
$(3, 9)$	$+$	$+$	$-$	$+$	$\frac{(+)(+)}{(-)(+)} = -$
$(9, \infty)$	$+$	$+$	$+$	$+$	$\frac{(+)(+)}{(+)(+)} = +$

Key insight: A fraction is negative when it has an odd number of negative factors in total.

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$f(x) < 0$ when $x \in (-5, -2) \cup (3, 9)$

In other words: $-5 < x < -2$ or $3 < x < 9$

We use open intervals because the function is undefined at $x = -2$ and $x = 9$, and equals zero (not negative) at $x = -5$ and $x = 3$.