Respuesta :
Answer:
(-6,2)
Step-by-step explanation:
[tex]f(x)<0[/tex]
We are given [tex]f(x)=x^2+4x-12[/tex].
So we need to solve:
[tex]x^2+4x-12<0[/tex]
First let's find when the expression on the left is 0.
[tex]x^2+4x-12=0[/tex].
The expression on the left is factorable.
Given the coefficient of [tex]x^2[/tex] is 1, all we have to do is see if there are two numbers that multiply to be -12 and add to be 4.
Those numbers are 6 and -2 so the factored form is:
[tex](x+6)(x-2)=0[/tex]
So the solutions can be found be solving the following linear equations:
[tex]x+6=0[/tex] and [tex]x-2=0[/tex].
First equation, you just need to subtract 6 on both sides. [tex]x=-6[/tex]
Second equation, you just need to add 2 on both sides. [tex]x=2[/tex]
So draw a number line with those on it and test the intervals they divide.
--------(-6)--------(2)----------
So we need to test three intervals.
*Plug in a number before -6 to see if that piece of the function is above the x-axis or below. We are looking for below since we want [tex]f(x)<0[/tex].
Let's plug in x=-8
[tex]f(x)=(x+6)(x-2)[/tex] or use [tex]f(x)=x^2+4x-12[/tex].
[tex]f(-8)=(-8+6)(-8-2)[/tex]
[tex]f(-8)=(-2)(-10)[/tex]
[tex]f(-8)=20[/tex] so this piece is above since 20 is positive and not negative.
So we will not include this interval in our answer.
*Plug in a number between -6 and 2 like 0.
[tex]f(x)=(x+6)(x-2)[/tex]
[tex]f(0)=(0+6)(0-2)[/tex]
[tex]f(0)=(6)(-2)[/tex]
[tex]f(0)=-12[/tex] so this piece is under since -12 is negative and not positive.
So our solution includes the interval (-6,2).
*Plug in a number after 2 like 5.
[tex]f(x)=(x+6)(x-2)[/tex]
[tex]f(5)=(5+6)(5-2)[/tex]
[tex]f(5)=(11)(3)[/tex]
[tex]f(5)=33[/tex] so this piece is above since 33 is positive and not negative.
The answer only includes the interval (-6,2).
