﻿ 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

#### Multiple Choice Question (MCQ)

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 5 months ago

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 5 months ago

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