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Math Help - Implied Domain

  1. #1
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    Implied Domain

    I have a test tomorrow at 10 am.. it's 9:30 right now. I really need help with understanding implied domain. my teacher never fully explained it, and now i'm panicing. He would like us to put it in interval notation, with teh sideways 8 looking things. PLEASE HELP!

    also, how to put a quadratic function into vertex form...
    i.e. f(x)= x^2+5x+6 and g(x)= x^2+3
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  2. #2
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    Quote Originally Posted by Mmarie View Post
    I have a test tomorrow at 10 am.. it's 9:30 right now. I really need help with understanding implied domain. my teacher never fully explained it, and now i'm panicing. He would like us to put it in interval notation, with teh sideways 8 looking things. PLEASE HELP!

    also, how to put a quadratic function into vertex form...
    i.e. f(x)= x^2+5x+6 and g(x)= x^2+3
    for f(x), you must complete the square. (can you continue?)

    g(x) is already in it's vertex form. unless you want to get technical and write it as g(x) = (x - 0)^2 + 3
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  3. #3
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    Thank you!

    well, this is an example and i'm totally confused as to wear my teacher got a 2... you'll understand when you see this:

    Finding the vertex by completing the square

    h(x)=x^2+9x+20
    =x^2+9x+(9/2)^2-(9/2)^2+20
    =(x+9/2)^2-81/4+20
    =(x+9/2)^2-81/4+80/4
    =(x+9/2)^2-1/4
    and it says the vertex is (-9/2, -1/4)

    ..i don't understand where the two that divides the 9 comes from. i'm totally lost!
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    Quote Originally Posted by Mmarie View Post
    Thank you!

    well, this is an example and i'm totally confused as to wear my teacher got a 2... you'll understand when you see this:

    Finding the vertex by completing the square

    h(x)=x^2+9x+20
    =x^2+9x+(9/2)^2-(9/2)^2+20
    =(x+9/2)^2-81/4+20
    =(x+9/2)^2-81/4+80/4
    =(x+9/2)^2-1/4
    and it says the vertex is (-9/2, -1/4)

    ..i don't understand where the two that divides the 9 comes from. i'm totally lost!
    it is a property of perfect squares that the lone constant is always the square of 1/2 the coefficient of x. since the coefficient of x is 9, to have a complete square, we need the constant to be (9/2)^2. since it was not there, we add it, but since we added something that was not there, we subtract it again right after. then we can transform the first three terms into a complete sqaure, and simplify the left overs to get what you see
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  5. #5
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    okay yes, i sort of understand. so what about a problem like..

    x^2+6x-7 and x^2+5x+6
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  6. #6
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    i just tried the first one.. x^2+6x+7

    the correct vertex is (-3,-16) but i got (-3,-2)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mmarie View Post
    i just tried the first one.. x^2+6x+7

    the correct vertex is (-3,-16) but i got (-3,-2)
    you posted x^2 + 6x - 7 last time. which is it?
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  8. #8
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    it's +7.. sorry for the confusion!
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  9. #9
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    Quote Originally Posted by Mmarie View Post
    it's +7.. sorry for the confusion!
    it is not (-3,-16)

    y = x^2 + 6x + 7

    \Rightarrow y = x^2 + 6x + 3^2 - 3^2 + 7

    \Rightarrow y = (x + 3)^2 - 9 + 7

    \Rightarrow y = (x + 3)^2 - 2

    thus, the vertex is (-3,-2)
    Attached Thumbnails Attached Thumbnails Implied Domain-parabola.jpg  
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  10. #10
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    also, is there anyway you can help me with three more things..

    the inverse of--

    f(x)=(x+2)/(x+9) and f(x)=(5)/(11x-6)


    evaluating functions..
    k(x)=3x^2-7 for k(2x^5)

    my answer was 6x^7-7, but the answer is 12x^10-7


    and finding the slope-intercept form when being given a point and the slope.
    for example.. m=-3 with the point (-1,2)


    PS. i really appreciate all of your help!
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mmarie View Post
    also, is there anyway you can help me with three more things..

    the inverse of--

    f(x)=(x+2)/(x+9) and f(x)=(5)/(11x-6)
    say for the first.

    f(x) = (x + 2)/(x + 9)

    Let y = f(x)

    so we have

    y = (x + 2)/(x + 9)

    for the inverse, switch x and y and solve for y. that will give you the inverse function

    that is, solve x = (y + 2)/(y + 9) for y


    the other is done in the same way


    evaluating functions..
    k(x)=3x^2-7 for k(2x^5)

    my answer was 6x^7-7, but the answer is 12x^10-7
    k(2x^5) means, you replace all the x's in k(x) with 2x^5

    so, k(2x^5) = 3(2x^5)^2 - 7

    now simplify, bearing in mind the laws of exponents

    and finding the slope-intercept form when being given a point and the slope.
    for example.. m=-3 with the point (-1,2)
    start with the point-slope form:

    the point slope form is: y - y_1 = m(x - x_1). where m is the slope, and (x_1,y_1) is a point the line passes through. if we expand the right hand side of this and solve for y, we get the slope intercept form: y = mx + b

    so, you want to solve y - 2 = -3(x + 1) for y
    Last edited by Jhevon; March 7th 2008 at 01:52 PM.
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