Multiple Choice Question (MCQ)

If the length of three sides of \triangle ABC is a,b,c, such that when x=-\dfrac{1}{2}, the min value of quadratic function y=(a+b)x^2+2cx-(a-b) is -\dfrac{b}{2}, then \triangle ABC is a(n)

  1. equilateral triangle

  2. ×

    isosceles triangle

  3. ×

    right triangle

  4. ×

    acute triangle

Collected in the board: Quadratic function

Steven Zheng posted 2 weeks ago

Answer

  1. For a standard quadratic function,

    y=ax² +bx+c

    the coordinates of the vertex of the curve is

    \Big( \dfrac{-b}{2a},\dfrac{4ac-b^2}{4a} \Big) (a, b,c is not related to the given question here).


    For given quadratic function,

    y=(a+b)x^2+2cx-(a-b)
    (1)

    the function for axis of symmetry is

    x = \dfrac{-2c}{2(a+b)} = - \dfrac{1}{2}

    Then,

    a+b=2c
    (2)

    Substitute (2) to (1)

    y=2cx^2+2cx-(a-b)

    =2c(x+\dfrac{1}{2} )^2-\dfrac{c}{2} -(a-b)

    Therefore

    y_{min} = -\dfrac{c}{2} -(a-b) =-\dfrac{b}{2}

    Then,

    c=3b-2a
    (3)

    Substituting (3) to (2) results in

    a+b= 6b-4a

    a=b
    (4)

    Substituting (4) to (2) yields

    a=c
    (5)

    Therefore

    a=b=c

    \triangle ABC is an equilateral triangle

Steven Zheng posted 1 week ago

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