# Thread: Need help w/ pre-calculus summer packet.

1. ## Need help w/ pre-calculus summer packet.

I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!

Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.

Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.

Factor completely.
3. x^2 + 12x + 36 - 9y^2
Stumped. I know I can factor out an x but not sure how that would help.

I appreciate any help. Look forward to hearing from people!

2. Originally Posted by amy1613
I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!

Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.

you should already know what y = x^2 + x - 12 looks like ... well, y = |x^2 + x - 12| is the same graph except that the part that is below the x-axis (negative) is reflected to the positive side.

y = x^2 + x - 12 , x < -4 or x > 3

y = -(x^2 + x - 12) , -4 < x < 3

Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.

x^2 - 16 > 0 ... note that x^2 - 16 = 0 at x = + 4 and x = -4

plot these two numbers on a number line ... it breaks the number line into three sections ...
x < -4 , -4 < x < 4 , and x > 4

take any number in each interval, and "test" it in the original inequality ... if it makes the inequality true, then all values of x in that interval make the inequality true.

you'll see that you forgot the interval (-∞ , -4)

Factor completely.
3. x^2 + 12x + 36 - 9y^2

help any?
...

3. yeah this helped a lot! thanks! so...a sign chart is just a number line graph?

4. Hello, amy1613!

Write the following absolute value expression as a piecewise expression:
. . $\displaystyle 1)\;\;f(x) \;=\;|x^2 + x - 12|$
There are two cases to consider: .$\displaystyle \begin{array}{cccc}(1) & x^2 + x - 12 \:\geq \;0 \\ (2) & x^2 + x - 12 \:<\:0 \end{array}$

$\displaystyle \text{Case }(1)\;\;x^2 + x - 12 \:\geq\:0\quad\Rightarrow\quad (x-3)(x+4) \:\geq \:0$

. . There are two ways:

. . . . $\displaystyle (a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$

. . . . $\displaystyle (b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$

. . Hence, if $\displaystyle x \geq 3\text{ or }x < -4$, then: .$\displaystyle f(x) \:=\:x^2+x-12$

$\displaystyle \text{Case }(2)\;\; x^2 + x - 12 \:<\: 0 \quad\Rightarrow\quad (x-3)(x+4) \:<\:0$

. . There are two ways:

. . . . $\displaystyle (a)\;\;\begin{Bmatrix}x-3 \:>\:0 & \Rightarrow & x \:>\:3 \\ x+4 \:< \:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x > 3 \text{ and }x < \text{-}4$ . . . impossible

. . . . $\displaystyle (b)\;\;\begin{Bmatrix}x-3 \:<\:0 & \Rightarrow & x \:<\:3 \\ x+4 \:> \:0 & \Rightarrow & x \:>\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad \text{-}4 \:<\:x \:<\: 3$

. . Hence, if $\displaystyle \text{-}4 < x < 3$, then: .$\displaystyle f(x) \:=\:-(x^2+x-12) \:=\:12 - x - x^2$

Therefore: . $\displaystyle f(x) \;=\;\begin{Bmatrix}x^2+x-12 && \text{if }x\leq \text{-}4 \text{ or }x \geq 3 \\ \\[-4mm] 12 -x-x^2 && \text{if }\text{-}4 < x < 3 \end{Bmatrix}$

5. Originally Posted by Soroban
. . There are two ways:

. . . . $\displaystyle (a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$

. . . . $\displaystyle (b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$

. . Hence, if $\displaystyle x \geq 3\text{ or }x < -4$, then: .$\displaystyle f(x) \:=\:x^2+x-12$

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Thanks for your response & explaination. I still can't quite understand why you picked x>-3 and not x>4. Could you go through that for me?