Determine the intervals for which $x ^ { 2 } - 4 x - 45$ is entirely negative and entirely positive.
A)negative: $( 5,9 )$ ; positive: $( - \infty , 5 ) \cup ( 9 , \infty )$
B)negative: $( - 5,9 )$ ; positive: $( - \infty , - 9 ) \cup ( 5 , \infty )$
C)negative: $( - 5,9 )$ ; positive: $( - \infty , - 5 ) \cup ( 9 , \infty )$
D)negative: $( 5,9 )$ ; positive: $( - \infty , - 9 ) \cup ( 5 , \infty )$
E)negative: $( - 9,5 )$ ; positive: $( - \infty , - 9 ) \cup ( 5 , \infty )$

Question

Quizplus · Accepted Answer

To determine the intervals for which the quadratic is entirely negative and entirely positive, we need to find the x-values that make the quadratic equal to 0, then test the intervals between and outside these x-values using test points. 
To find the x-values that make the quadratic equal to 0, we can factor as:
$$x^2 - 4x - 45 = (x-9)(x+5) = 0$$
So the zeros are at x=9 and x=-5. 
Next, we can test the intervals (-∞, -5), (-5, 9), and (9, ∞) using a test point from each interval. For example, we can choose x=-6 for interval (-∞, -5). Plugging in x=-6 into the quadratic gives:
$$(-6)^2 - 4(-6) - 45 = 51 > 0$$
Since this result is positive, the quadratic is positive for all x-values in the interval (-∞, -5). 
We can similarly test the other two intervals. Using x=0 for interval (-5, 9), we get:
$$0^2 - 4(0) - 45 = -45 < 0$$
Since this result is negative, the quadratic is negative for all x-values in the interval (-5, 9). 
Using x=10 for interval (9, ∞), we get:
$$10^2 - 4(10) - 45 = 11 > 0$$
Since this result is positive, the quadratic is positive for all x-values in the interval (9, ∞). 
Therefore, the correct answer is C with negative: $(-5,9)$ and positive: $(-\infty,-5) \cup (9,\infty)$.

Determine the Intervals for Which x2−4x−45x ^ { 2 } - 4 x - 45x2−4x−45

Determine the Intervals for Which $x ^ { 2 } - 4 x - 45$