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Thread: VCE Maths questions help

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    VCE Maths questions help

    1. How to find the turning point and type of turning point in the equation y = (2/3)x^4 + 1/3
    2. how to find equation of axis of symmetry of y = (2/3)x^4 + 1/3

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    1. How to find the turning point and type of turning point in the equation y = (2/3)x^4 + 1/3
    2. how to find equation of axis of symmetry of y = (2/3)x^4 + 1/3
    1) Turning point at 0, local minimum.

    2) Axis of symmetry is the y-axis.

    Method used: inspection.

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    Re: maths questions help

    2) My book said that the axis of symmetry is x = 0. Maybe my book's wrong?

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    2) My book said that the axis of symmetry is x = 0. Maybe my book's wrong?

    the y axis is x=0!!!
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    Re: maths questions help

    Quote Originally Posted by disturb_equilibrium View Post
    the y axis is x=0!!!
    Oh, right, I get it now, was just a bit confused before.

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    Re: maths questions help

    What's the difference between the turning point and the stationary point of inflection?

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    What's the difference between the turning point and the stationary point of inflection?
    A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point (i.e. gradient =0) However with turning points the concavity remains the same. In a stationary point of inflexion the gradient is 0 but the concavity changes, thus not changing from an increasing to a decreasing function or visa-versa.
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    Re: maths questions help

    Quote Originally Posted by Rathin View Post
    A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point (i.e. gradient =0) However with turning points the concavity remains the same. In a stationary point of inflexion the gradient is 0 but the concavity changes, thus not changing from an increasing to a decreasing function or visa-versa.
    So for x^2, x^4, x^6, x^8, etc, there's a turning point, and for x^3, x^5, x^7, x^9, etc there's a stationary point of inflection?

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    So for x^2, x^4, x^6, x^8, etc, there's a turning point, and for x^3, x^5, x^7, x^9, etc there's a stationary point of inflection?
    Correct!

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    Re: maths questions help

    Quote Originally Posted by InteGrand View Post
    Correct!
    Thanks

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    Re: maths questions help

    How hard it to get raw 40 in vce maths methods?

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    How hard it to get raw 40 in vce maths methods?
    Very few here are familiar with the VCE. But I find the VCE Maths curriculum modern and progressive.
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    Re: maths questions help

    Find the values of m if (2m-3)x^2 + (5m-1)x + (3m-2) = 0 has 2 solutions

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    Find the values of m if (2m-3)x^2 + (5m-1)x + (3m-2) = 0 has 2 solutions


    Remember, if the given quadratic equation has two distinct (real) solutions for x, then its discriminant must be greater than zero.

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    Re: maths questions help

    Quote Originally Posted by InteGrand View Post


    Remember, if the given quadratic equation has two distinct (real) solutions for x, then its discriminant must be greater than zero.
    I ended up getting 25m^2 - 10m + 1 - 4(6m^2 - 13m + 6)
    = 25m^2 - 10m + 1 - 24m^2 + 52m - 24
    = m^2 + 42m - 23 > 0

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    I ended up getting 25m^2 - 10m + 1 - 4(6m^2 - 13m + 6)
    = 25m^2 - 10m + 1 - 24m^2 + 52m - 24
    = m^2 + 42m - 23 > 0

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    Re: maths questions help

    A piece of wire 12 cm long is cut into two pieces. One piece is used to form a square shape and the other a rectangle shape in which the length is twice the width.
    a. If x cm is the side length of the square, write down the dimensions of the rectangle in terms of x
    b. formulate a rule for A, the combined area of the square and rectangle in cm^2, in terms of x.
    c. determine the lengths of the two pieces if the sum of the areas is to be a minimum.

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    Re: maths questions help







    Last edited by kawaiipotato; 29 Apr 2017 at 9:45 PM.
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    Re: maths questions help

    Is it normal to find maths hard even after tution? I'm in this scenario

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    Re: maths questions help

    The graph of y = x^4 - 2x - 12 has 2 x-intercepts
    a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
    b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
    c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

    I get how to do part a and b, but i'm finding part c extremely challenging and confusing.

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    The graph of y = x^4 - 2x - 12 has 2 x-intercepts
    a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
    b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
    c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

    I get how to do part a and b, but i'm finding part c extremely challenging and confusing.
    If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).

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    Re: maths questions help

    I read my textbook but I still don't understand the bisection methods.

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    I read my textbook but I still don't understand the bisection methods.
    Are there any examples in the textbook using bisection method? See this page if not (there's an example in it): https://en.wikipedia.org/wiki/Bisection_method .
    kawaiipotato likes this.

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    Re: maths questions help

    Can someone please help?
    The graph of y = x^4 - 2x - 12 has 2 x-intercepts
    a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
    b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
    c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

    I get how to do part a and b, but i'm finding part c extremely challenging and confusing.

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    Re: maths questions help

    Quote Originally Posted by boredsatan View Post
    Can someone please help?
    The graph of y = x^4 - 2x - 12 has 2 x-intercepts
    a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
    b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
    c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

    I get how to do part a and b, but i'm finding part c extremely challenging and confusing.
    Quote Originally Posted by InteGrand View Post
    If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).
    .

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