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 $$ riangle ACB$$, such that when $$x =1$$, the minimum value of the function is $$-\dfrac{8}{5}b $$, then $$ riangle ACB$$ is a(n)

Question

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 $$	riangle ACB$$, such that when $$x =1$$, the minimum value of the function is $$-\dfrac{8}{5}b  $$, then $$	riangle ACB$$ is a(n)

Uniteasy Admin · Accepted Answer

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
$$y = (a-\dfrac{b}{2}  )x^2-cx-a-\dfrac{b}{2} $$n
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
$$c=\dfrac{3}{5}b $$n
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
$$a = \dfrac{4}{5}b $$n
Then,
$$c:b:a = \dfrac{3}{5}b:b:\dfrac{4}{5}b = 3:5:4$$
Therefore, $$	riangle ACB$$ is a right triangle.

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