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Thread: Domain and Range

  1. #1
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    Domain and Range

    Find the domain and range for each quadratic function.

    a. y = x^2 – 6x +5

    b. y = 3x^2 + 2x +9
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pashah
    Find the domain and range for each quadratic function.

    a. y = x^2 6x +5

    b. y = 3x^2 + 2x +9
    The RHS of both of these are defined for all of $\displaystyle \mathbb{R}$, so
    their domains are both the set of all real numbers.

    Because these both represent parabolas which go to $\displaystyle \infty$ as $\displaystyle x$
    goes to $\displaystyle \pm \infty$, they can take all values greater than teir minimums.

    The minimum of the first occurs when $\displaystyle 2x-6=0$, which is at
    $\displaystyle x=3$, where $\displaystyle y=-4$, so for a. the range is $\displaystyle y \ge 4$.

    The minimum of the second occurs when $\displaystyle 6x+2=0$, which is at
    $\displaystyle x=-1/3$, where $\displaystyle y=26/3$, so for a. the range is $\displaystyle y \ge 26/3$.

    RonL
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  3. #3
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    Quote Originally Posted by pashah
    Find the domain and range for each quadratic function.

    a. y = x^2 6x +5
    I do this slight diffrently.
    Rewrite as,
    $\displaystyle x^2-6x+(5-y)=0$
    Now, you need to find all $\displaystyle y$ such as this quadradic equation has a real solution. That happens when, the discrimanant is non-negative.
    $\displaystyle (-6)^2-4(1)(5-y)\geq 0$
    Simplify,
    $\displaystyle 16+4y\geq 0$
    Thus,
    $\displaystyle 4y\geq -16$
    Thus,
    $\displaystyle y\geq -4$
    b. y = 3x^2 + 2x +9
    Proceede the same,
    $\displaystyle 3x^2+2x+(9-y)=0$
    Work with discrimanant,
    $\displaystyle (2)^2-4(3)(9-y)\geq 0$
    $\displaystyle -104+12y\geq0$
    $\displaystyle 12y\geq 104$
    $\displaystyle y\geq 8.\bar 6$
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