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Thread: Polynomial Func. and Eq.'s

  1. #1
    Raj
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    Polynomial Func. and Eq.'s

    If anyone could provide the answer so i can compare with mine, or tell me whats wrong with my answer that would be great.

    1. Solve Algebraically

    A) $\displaystyle 2x^2+5x-63>0$

    My Answer: $\displaystyle {-7>x<9/2}$

    B) $\displaystyle 20x^3-3x^2-17x+6 \leq 0$

    My Answer: $\displaystyle {-1>x,2/5<x<3/4}$

    C) $\displaystyle \frac{x+1}{x-3} - \frac{2x^2+3x}{4x^2-9} = \frac{5}{2x^2-9x+9}$

    My Answer: $\displaystyle
    {x=2,-4}
    $




    2. Sketch the graph of $\displaystyle f(x)=\frac{x+4}{x-5}$, Determine all intercepts, the domain and the range, and asymptotes.
    Sketch the graph of $\displaystyle g(x)=3x-8$ on the same graph as f(x). Determine where $\displaystyle f(x) \geq g(x)$.

    Intercepts: $\displaystyle y=-4/5$, $\displaystyle x=-4$
    Domain and Range: Don't know how to write it on here
    Vertical Asymptote: $\displaystyle x=5$
    Horizontal Asymptote: $\displaystyle y=1$

    I was able to graph it, I just need help on determining where $\displaystyle f(x) \geq g(x)$.
    Last edited by Raj; Nov 9th 2007 at 07:37 AM.
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  2. #2
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    2x^2 + 5x - 63

    solving this, you get x= -7 and 9/2

    2x^2 + 5x - 63 > 0

    check if it holds true when x is smaller than -7, between -7 and 9/2, and greater than 9/2.

    2(-9999)^2 + 5(-9999) - 63 > 0
    huge positive number - smaller numbers > 0
    positive > 0 (check!)

    2(0)^2 + 5(0) - 63 > 0
    -63 > 0 (.. nope!)

    2(9999)^2 + 5(9999) - 63 > 0
    huge positive number + smaller number - even smaller number > 0
    positive > 0 (check!)

    Therefore, x holds true when x < -7 and x > 9/2
    (-infinite, -7) U (9/2, infinite)
    (Note: not inclusive since it isn't 2x^2 + 5x - 63 >= 0)

    --same concept for b-- btw, this is just one way to solve for x.
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  3. #3
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    Quote Originally Posted by Raj View Post
    My Answer: $\displaystyle {-7>x<9/2}$
    Nice try on the notation shorthand, but never write anything like that again.

    Trust me on this. Very ugly. If I were grading it, I would mark it wrong, at least partially.
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  4. #4
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    Hello, Raj!

    1. Solve algebraically

    $\displaystyle A)\;\;2x^2+5x-63\:>\:0$

    The graph of $\displaystyle y \:=\:2x^2+5x-63$ is an up-opening parabola.
    . . We want to know when the graph is above the x-axis.
    The function is positive to the left and the right of its x-intercepts.

    x-intercepts: .$\displaystyle 2x^2+5x-63 \:=\:0\quad\Rightarrow\quad (x + 7)(2x - 9)\:=\:0\quad\Rightarrow\quad x \:=\:-7,\:\frac{9}{2}$
    Code:
                      |
         *            |        *
                      |
        --o-----------+-------o--
            *         |     *
              *       |   *
                    * |
                      |
    Answer: .$\displaystyle \boxed{x < -7\:\text{ or }\:x > \frac{9}{2}}$


    $\displaystyle B)\;\;20x^3-3x^2-17x+6 \:\leq \:0$

    We have the cubic function: .$\displaystyle y \:=\:20x^3 - 3x^2 - 17x + 6$
    . . We want know when this cubic is below the x-axis.

    It factors: .$\displaystyle (x + 1)(5x-2)(4x-3) \:=\;0\quad\Rightarrow\quad x \:=\:-1,\:\frac{2}{5},\:\frac{3}{4}$
    Code:
                    |
                  * |           *
               *    |*
          ----o-----+-o--------o---
                    |
             *      |  *     *
                    |     *
                    |
    Answer: .$\displaystyle \boxed{x < -1,\;\frac{2}{5} < x < \frac{3}{4}}$
    Your answers are correct!


    $\displaystyle C)\;\;\frac{x+1}{x-3} - \frac{2x^2+3x}{4x^2-9} \:= \:\frac{5}{2x^2-9x+9}$

    We have: .$\displaystyle \frac{x+1}{x-3} - \frac{x(2x+3)}{(2x -3)(2x+3)} \:=\;\frac{5}{(x-3)(2x-3)}$

    . . $\displaystyle = \;\frac{x+1}{x-3} - \frac{x}{2x-3} \;=\;\frac{5}{(x-3)(2x-3)} $


    Multiply by the LCD, $\displaystyle (x-3)(2x-3)$
    . . $\displaystyle (x+1)(2x-3) - x(x-3) \:=\:5\quad\Rightarrow\quad x^2 + 2x - 8 \:=\:0\quad\Rightarrow\quad(x+4)(x-2) \:=\:0$

    Therefore: .$\displaystyle \boxed{x \;=\;-4,\:2}$

    You were right!
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