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Math Help - Graphs of basic functions & Remainder-Factor theorem

  1. #1
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    Graphs of basic functions & Remainder-Factor theorem

    Hello! I have a few questions regarding the graphs of basic functions (eg. y = x^3). Please provide explanations to the answers if possible! Thank You very much!

    Q1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

    Q2) It is not possible to construct a cubic equation that describes a graph that does not intersect with the x-axis. Why not? State and explain using the Remainder and Factor Theorem.

    Q3) Given that the 2 quadratic equations:
    x^2 -nx -36 = 0
    n + 6x - x^2 = 0
    have a common root x = a for certain values of the real constant n, show that a^3 - 7(a^2) + 36 = 0

    Hence find the possible values of a, and the corresponding values of n.

    ---
    Thank you very much!
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  2. #2
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    Hello, Tangera!

    3) Given that the two quadratic equations: . \begin{array}{ccc}x^2 -nx -36 & = & 0 \\ x^2-6x-n &=&0\end{array}

    have a common root x = a for certain values of the real constant n,

    (a) Show that: .  a^3 - 7a^2 + 36 \:= \:0

    Since a is a root of both equations, we have:

    . . \begin{array}{cccccccc}a^2-an-36 &=&0 & \Longrightarrow & n & = &\dfrac{a^2-36}{a} & {\color{blue}[1]} \\ \\ a^2-6a-n &=&0 & \Longrightarrow & n &=&a^2-6a & {\color{blue}[2]} \end{array}


    Equate [1] and [2]: . \frac{a^2-36}{a} \:=\:a^2-6a\quad\Longrightarrow\quad\boxed{ a^3 - 7a^2 + 36 \:=\:0}




    \text{(b) Hence, find the possible values of }a\text{ and the corresponding values of }n.

    The cubic factors: . (a-3)(a-6)(a+2) \:=\:0

    . . and has roots: . a \;=\;3,\:6,\:-2


    Substitute into [2]: . n \;=\;-9,\:0,\:16


    Therefore: . \boxed{(a,\,n) \;=\;(3,\,\text{-}9),\;(6,\,0),\;(\text{-}2,\,16)}

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  3. #3
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    I see! Thanks for explaining!
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  4. #4
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    For your first question,

    For a graph to have a stationary point, dy/dx must be 0 at some point.
    For a cubic graph, dy/dx is always a quadratic expression.
    When can't a quadratic expression = 0? When the discriminant is < 0.
    So, just find yourself a b^2-4ac that's < 0, then from there construct a quadratic expression. Then just integrate it for the cubic equation.

    Hope I helped.
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    Quote Originally Posted by stonetoad View Post
    For your first question,

    For a graph to have a stationary point, dy/dx must be 0 at some point.
    For a cubic graph, dy/dx is always a quadratic expression.
    When can't a quadratic expression = 0? When the discriminant is < 0.
    So, just find yourself a b^2-4ac that's < 0, then from there construct a quadratic expression. Then just integrate it for the cubic equation.

    Hope I helped.
    Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.
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    Quote Originally Posted by angel.white View Post
    Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.
    Yes, it's possible. For example, let's say the discriminant is (2^2)-4(3)(3). So the quadratic expression that results in this discriminant is x^3 + 2x +3. Then you integrate it to get the cubic expression x^3 + x^2 + 3x. It all makes sense, doesn't it?
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  7. #7
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    Quote Originally Posted by stonetoad View Post
    Yes, it's possible. For example, let's say the discriminant is (2^2)-4(3)(3). So the quadratic expression that results in this discriminant is x^3 + 2x +3. Then you integrate it to get the cubic expression x^3 + x^2 + 3x. It all makes sense, doesn't it?
    Oh, gotcha.
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    Quote Originally Posted by Tangera View Post
    Hello! I have a few questions regarding the graphs of basic functions (eg. y = x^3). Please provide explanations to the answers if possible! Thank You very much!

    Q1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

    Q2) It is not possible to construct a cubic equation that describes a graph that does not intersect with the x-axis. Why not? State and explain using the Remainder and Factor Theorem.

    Q3) Given that the 2 quadratic equations:
    x^2 -nx -36 = 0
    n + 6x - x^2 = 0
    have a common root x = a for certain values of the real constant n, show that a^3 - 7(a^2) + 36 = 0

    Hence find the possible values of a, and the corresponding values of n.

    ---
    Thank you very much!
    So to summarise:

    Questions 1 and 2 have been addressed here. Soroban has addressed question 3.
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  9. #9
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    Quote Originally Posted by angel.white View Post
    Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.
    All the theory (in the form of questions and solutions) is given in the attachment found here.
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  10. #10
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    Hello! Many thanks to everyone who pondered over the questions and helped answer them! I've learnt quite a bit from your answers and questions. Thank you very much!
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  11. #11
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    Quote Originally Posted by Tangera View Post
    Hello! Many thanks to everyone who pondered over the questions and helped answer them! I've learnt quite a bit from your answers and questions. Thank you very much!
    And on behalf of the forum, thankyou for your lovely manners.
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