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Math Help - Need help w/ pre-calculus summer packet.

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by amy1613 View Post
    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

    start with this ... (x + 6)^2 - (3y)^2

    help any?
    ...
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  3. #3
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    yeah this helped a lot! thanks! so...a sign chart is just a number line graph?
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  4. #4
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    Hello, amy1613!

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


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

    . . There are two ways:

    . . . . (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

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



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

    . . There are two ways:

    . . . . (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

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


    Therefore: . 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}


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  5. #5
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    Quote Originally Posted by Soroban View Post
    . . There are two ways:

    . . . . (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

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

    [/size]
    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?
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