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Thread: VCE Maths questions help

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    Re: VCE Maths questions help

    Given the graphs of g(x) = x^2 + 3x + 2 and 2y + mx + 8 = 0. Find m such that the graphs have
    one intersection

    no intersection

    Would I need to set the equations equal to each other for both?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Given the graphs of g(x) = x^2 + 3x + 2 and 2y + mx + 8 = 0. Find m such that the graphs have
    one intersection

    no intersection

    Would I need to set the equations equal to each other for both?
    Essentially.

    (Make sure to first express y in terms of x for the 2y + mx + 8 = 0.)

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    Essentially.

    (Make sure to first express y in terms of x for the 2y + mx + 8 = 0.)
    Thanks for claryfying
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    could it be (2,infinity), even though the graph never touches 2?
    That's the range in the "old" notation. Maybe for VCE, like for the IB, you use ]2, infinity symbol [, i.e.

    Last edited by Drongoski; 15 May 2017 at 11:00 PM.
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    Re: VCE Maths questions help

    Quote Originally Posted by Drongoski View Post
    That's the range in the "old" notation. Maybe for VCE, like for the IB, you use ]2, infinity symbol [, i.e.

    Fairly sure they use the parentheses in VCE.

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    Re: VCE Maths questions help

    Quote Originally Posted by leehuan View Post
    Fairly sure they use the parentheses in VCE.
    Yes, parentheses and brackets are used in vce
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    could it be (2,infinity), even though the graph never touches 2?
    So would the range of y = 2/(x+2) + 2 be R {2} ?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    So would the range of y = 2/(x+2) + 2 be R {2} ?
    ℝ \ {2} (You forgot a backslash.)

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    ℝ \ {2} (You forgot a backslash.)
    Is there a reason why for y = 2/(x+2)^2 + 2, the range is (2, infinity),
    but for y = 2/(x+2) + 2, the range is R\ {2}
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Is there a reason why for y = 2/(x+2)^2 + 2, the range is (2, infinity),
    but for y = 2/(x+2) + 2, the range is R\ {2}
    Yes. In the first one, since there is a 2/(x+2)^2, this can only take on positive values (and moreover can take on any positive value), because of the square in the denominator (and positive constant numerator). This causes the range to become (2, infinity).

    With the second one, the 2/(x+2) can take on any non-zero real value (including negative ones), and can't take the value 0. This makes the range be what you wrote (ℝ \ {2}).

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    Yes. In the first one, since there is a 2/(x+2)^2, this can only take on positive values (and moreover can take on any positive value), because of the square in the denominator (and positive constant numerator). This causes the range to become (2, infinity).

    With the second one, the 2/(x+2) can take on any non-zero real value (including negative ones), and can't take the value 0. This makes the range be what you wrote (ℝ \ {2}).
    Although the domain for both would be R\{2}?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Although the domain for both would be R\{2}?
    Correct!
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    Re: VCE Maths questions help

    Bump!
    How would you express y = 2x^2 - 8x + 12 in turning point form
    Is this working right?
    2(x^2 - 4x + 6)
    x^2 - 4x + 4 - 4 + 6
    (x-2)^2 + 2
    2(x-2)^2+2
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Bump!
    How would you express y = 2x^2 - 8x + 12 in turning point form
    Is this working right?
    2(x^2 - 4x + 6)
    x^2 - 4x + 4 - 4 + 6
    (x-2)^2 + 2
    2(x-2)^2+2
    That's correct. So

    y - 2 = 2 (x - 2)^2

    So the turning point (equivalently the vertex of the parabola) is (2,2))
    Last edited by Drongoski; 16 May 2017 at 9:21 PM.
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    Re: VCE Maths questions help

    y = 0.5x + 5
    3y = √(3) x + 15
    are these the same?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    y = 0.5x + 5
    3y = √(3) x + 15
    are these the same?
    No

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    Re: VCE Maths questions help

    find the equation of a line which makes an angle of 30 degrees with the positive x axis and cuts the y axis at 5
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    find the equation of a line which makes an angle of 30 degrees with the positive x axis and cuts the y axis at 5
    Anyone?
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    Re: VCE Maths questions help

    Then this line makes an angle of 30 deg with the positive x-axis. Its gradient is tan(30 deg) = 1/sqrt(3)
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    Re: VCE Maths questions help

    Quote Originally Posted by Drongoski View Post
    Then this line makes an angle of 30 deg with the positive x-axis. Its gradient is tan(30 deg) = 1/sqrt(3)
    is the answer 1/sqrt(3) x + 5 ?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    is the answer 1/sqrt(3) x + 5 ?
    That isn't an equation.

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    Re: VCE Maths questions help

    Quote Originally Posted by 1729 View Post
    That isn't an equation.
    y = 1/sqrt(3) x + 5 ?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    y = 1/sqrt(3) x + 5 ?
    Correct!

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    No
    y = 1/sqrt(3) x + 5
    3y = √(3) x + 15
    are these the same?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    y = 1/sqrt(3) x + 5
    3y = √(3) x + 15
    are these the same?
    Yes

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