Question

l)determine the interval for where $f(x)>0$ (*hint: you need to factor)

1. $f(x)=x^{2}-4x - 12$
2. $f(x)=x^{2}+1x - 20$

Explanation:

Step1: Factor the quadratic function

For $f(x)=x^{2}-4x - 12$, we find two numbers that multiply to $- 12$ and add up to $-4$. These numbers are $-6$ and $2$. So $f(x)=(x - 6)(x+2)$.

Step2: Find the roots

Set $f(x)=0$, then $(x - 6)(x + 2)=0$. By the zero - product property, $x-6=0$ or $x + 2=0$. So the roots are $x=6$ and $x=-2$.

Step3: Test intervals

We have three intervals to test: $(-\infty,-2)$, $(-2,6)$ and $(6,\infty)$.
For $x<-2$, let $x=-3$. Then $f(-3)=(-3 - 6)(-3 + 2)=(-9)\times(-1)=9>0$.
For $-2For $x>6$, let $x=7$. Then $f(7)=(7 - 6)(7 + 2)=9>0$.
So the interval where $f(x)>0$ is $(-\infty,-2)\cup(6,\infty)$.

For $f(x)=x^{2}+x - 20$, we find two numbers that multiply to $-20$ and add up to $1$. These numbers are $5$ and $-4$. So $f(x)=(x + 5)(x-4)$.
Set $f(x)=0$, then $(x + 5)(x - 4)=0$. By the zero - product property, $x+5=0$ or $x - 4=0$. So the roots are $x=-5$ and $x = 4$.
We have three intervals to test: $(-\infty,-5)$, $(-5,4)$ and $(4,\infty)$.
For $x<-5$, let $x=-6$. Then $f(-6)=(-6 + 5)(-6 - 4)=(-1)\times(-10)=10>0$.
For $-5For $x>4$, let $x=5$. Then $f(5)=(5 + 5)(5 - 4)=10>0$.
So the interval where $f(x)>0$ is $(-\infty,-5)\cup(4,\infty)$.

Answer:

1. $(-\infty,-2)\cup(6,\infty)$
2. $(-\infty,-5)\cup(4,\infty)$