Question

For the function f(x) = x2 + 4x - 12 solve the following. f(x) < 0

Answer 1

Answer:

(-6,2)

Step-by-step explanation:

$f(x)$

We are given $f(x)=x^2+4x-12$.

So we need to solve:

$x^2+4x-12$

First let's find when the expression on the left is 0.

$x^2+4x-12=0$.

The expression on the left is factorable.

Given the coefficient of $x^2$ 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:

$(x+6)(x-2)=0$

So the solutions can be found be solving the following linear equations:

$x+6=0$ and $x-2=0$.

First equation, you just need to subtract 6 on both sides. $x=-6$

Second equation, you just need to add 2 on both sides. $x=2$

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 $f(x)$.

Let's plug in x=-8

$f(x)=(x+6)(x-2)$ or use $f(x)=x^2+4x-12$.

$f(-8)=(-8+6)(-8-2)$

$f(-8)=(-2)(-10)$

$f(-8)=20$ 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.

$f(x)=(x+6)(x-2)$

$f(0)=(0+6)(0-2)$

$f(0)=(6)(-2)$

$f(0)=-12$ 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.

$f(x)=(x+6)(x-2)$

$f(5)=(5+6)(5-2)$

$f(5)=(11)(3)$

$f(5)=33$ so this piece is above since 33 is positive and not negative.

The answer only includes the interval (-6,2).