Question

for the quadratic function f(x)=-x² - 2x, answer parts (a) through (f). (type your answer in interval notation.) (f) determine where f(x)>0 and where f(x)<0. select the correct choice below and fill in the answer box(es) within your choice. (type your answer in interval notation. use integers or fractions for any numbers in the expression.) a. f(x)<0 on , and f(x) is never positive b. f(x)>0 on , and f(x) is never negative c. f(x)>0 on , and f(x)<0 on

Explanation:

Step1: Set $f(x)=0$

$-x^{2}-2x = 0$
Factor out $-x$: $-x(x + 2)=0$

Step2: Solve for $x$

Set each factor equal to zero:
$-x=0$ gives $x = 0$; $x+2=0$ gives $x=-2$

Step3: Analyze the sign of the quadratic function

The quadratic function $y=-x^{2}-2x$ is a parabola opening downwards (because the coefficient of $x^{2}$ is negative, $a=-1<0$).
We test the intervals $(-\infty,-2)$, $(-2,0)$ and $(0,\infty)$
For $x<-2$, let $x=-3$, then $f(-3)=-(-3)^{2}-2\times(-3)=-9 + 6=-3<0$
For $-20$
For $x>0$, let $x = 1$, then $f(1)=-1^{2}-2\times1=-1-2=-3<0$

Answer:

C. $f(x)>0$ on $(-2,0)$ and $f(x)<0$ on $(-\infty,-2)\cup(0,\infty)$