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Math Help - axis of symmerty, etc

  1. #1
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    axis of symmerty, etc

    F(x)=-2X^2+3X+5

    axis of symmetry
    vertex
    x and y interceps
    domain and range
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  2. #2
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    Quote Originally Posted by ashleeyy- View Post
    F(x)=-2X^2+3X+5

    axis of symmetry
    vertex
    x and y interceps
    domain and range
    To learn how to find intercepts, try here. You'll also need to know how to solve quadatic equations; the Quadratic Formula may prove helpful.

    To learn how to find domains and ranges, try here. The domain is pretty obvious, since the function is a polynomial. To find the specific range here, you'll need to complete the square to convert the quadratic to "vertex" form. Then copy down the vertex. Use the fact that this quadratic is negative to determine whether the vertex gives the maximal value (and thus the upper limit of the range) or the minimal value (and thus the lower limit of the range).

    The axis of symmetry is the vertical line through the vertex.

    If you get stuck, please reply showing what you have tried so far. Thank you!
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  3. #3
    RRH
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    Quote Originally Posted by ashleeyy- View Post
    F(x)=-2X^2+3X+5

    axis of symmetry
    vertex
    x and y interceps
    domain and range
    Please try using Stapel sources before using my results. I am trying to practice using latex and it took me approx 45 minutes to type this up (yes I am slow) and I do not want to lose my work; therefore, I am going to post it, but you should try to do the work yourself first.

    Vertex = ( \frac{3}{4} , \frac{49}{8})

    x-int = -1 and \frac{5}{2}

    y-int = 5

    domain is (-inf,+inf)

    range is (-inf, \frac{49}{8}]

    See explanation below to determine how I came to these results.

    Completing the Square

    <br />
f(x) = -2x^2+3x+5<br />

    When completing the square the value of "a" should not be any number other than 1; therefore, factor out the -2

    <br />
-2(x^2-\frac{3}{2}x)+5<br />

    Now divide -\frac{3}{2} by 2 and square the result. The equation should now look like this.

    <br />
-2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16})+5<br />

    To move the  -\frac {9}{16} out of the parenthesis you have to multiply it by -2

    <br />
-2(x^2-\frac{3}{2}x+\frac{9}{16})+5+\frac{9}{8}<br />

    Factor within the parenthesis and clean up outside the parenthesis. The vector format of your function should look like below

    <br />
-2(x-\frac{3}{4})^2+\frac{49}{8}<br />

    To determine the y-int take the original function f(x) = -2x^2+3x+5 substitute 0 for x and you should get the y-int of 5

    To get the x-int take the vertex format of the function -2(x-\frac{3}{4})^2+\frac{49}{8} and substitute 0 for y

    -2(x-\frac{3}{4})^2+\frac{49}{8}=0

    -2(x-\frac{3}{4})^2=-\frac{49}{8}

    Divide both sides by -2

    (x-\frac{3}{4})^2=\frac{49}{16}

    \sqrt(x-\frac{3}{4})^2=\sqrt\frac{49}{16}

    x-\frac{3}{4}=\pm\frac{7}{4}

    x=\frac{3}{4}\pm\frac{7}{4}

    the x-int = -1 , \frac{5}{2}
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  4. #4
    Senior Member pacman's Avatar
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    f(x) = -2x^2 + 3x + 5 = y,

    completing the square;

    x^2 - (3/2)x = (-1/2)y + 5/2,

    x^2 - (3/2)x + (3/4)^2 = (-1/2)y + 5/2 + (3/4)^2,

    (x - 3/4)^2 = (-1/2)y + 5/2 + 9/16 = (-1/2)y + 49/16

    (x - 3/4)^2 = (-1/2)(y - 49/8), this is the standard form of a parabola facing downward since 4p = -1/2.

    axis of symmetry is vertical, with equation equal to x = 3/4

    vertex: (h,k) = (3/4, 49/8)

    y-intercept? set x = 0, y = 5.

    x-intercept? set y = 0, solve for x: that is (x - 3/4)^2 = (-1/2)(y - 49/8)

    (x - 3/4)^2 = (-1/2)(0 - 49/8) = 49/16

    (x - 3/4)^2 - (7/4)^2 = 0, by Zero factor Theorem,

    x - 3/4 - 7/4 = 0, then x = 10/4 = 5/2, and

    x - 3/4 + 7/4 = 0, then x = -4/4 = -1. Thus x-intercepts are,{-1 and 5/2}.

    Domain and Range? . . . . look at the graph.

    Domain: set of real numbers, - inf to + inf,

    Range: y is less than or equal to 49/8.



    Attached Thumbnails Attached Thumbnails axis of symmerty, etc-par.gif   axis of symmerty, etc-par-2.gif  
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  5. #5
    RRH
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    I like the graphs .. how do you do them?
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  6. #6
    Senior Member pacman's Avatar
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    there are so many programs for that . . . . most of the time, i use this wolframalpha
    Wolfram|Alpha
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  7. #7
    RRH
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    Quote Originally Posted by pacman View Post
    there are so many programs for that . . . . most of the time, i use this wolframalpha
    Wolfram|Alpha

    Thank you
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