Multiple Choice Question (MCQ)

Suppose a quadratic function

y = (a-\dfrac{b}{2} )x^2-cx-a-\dfrac{b}{2}

is constructed using three sides a,b,c of \triangle ACB, such that when x =1, the minimum value of the function is -\dfrac{8}{5}b , then \triangle ACB is a(n)

  1. ×

    isosceles triangle

  2. ×

    acute triangle

  3. ×

    obtuse triangle

  4. right 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 minimum value happens at the vertex of the curve with coordinates

    \Big( \dfrac{-b}{2a},\dfrac{4ac-b^2}{4a} \Big)

    For the given function

    y = (a-\dfrac{b}{2} )x^2-cx-a-\dfrac{b}{2}
    (1)

    Since y_{min} = -\dfrac{8}{5}b , substitute x=1 into (1),

    (a-\dfrac{b}{2} ) -c -a-\dfrac{b}{2} = 0

    Simplify the equation, we get

    c=\dfrac{3}{5}b
    (2)

    Since x =1, we have the following equation

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

    Substitue (2), simplify and solve for a, we get

    a = \dfrac{4}{5}b
    (3)

    Then,

    c:b:a = \dfrac{3}{5}b:b:\dfrac{4}{5}b = 3:5:4

    Therefore, \triangle ACB is a right triangle.


Steven Zheng posted 2 weeks ago

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