If the fraction function $$y = \dfrac{ax^2+bx+6}{x^2+2}$$ has minimum value 2 and maximum value 6, find the value of $$a,b$$

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If the fraction function $$y = \dfrac{ax^2+bx+6}{x^2+2}$$  has minimum value 2 and maximum value 6, find the value of $$a,b$$

Uniteasy Admin · Accepted Answer

Rearrange the function
$$ax^2+bx+6 = y(x^2+2)$$
then,
$$(y-a)x²-bx+2y-6=0$$
which implies $$y
e a$$, or it will be possible to have min or max values.
For a quadratic equation to have real solutions, 
the discriminant must be greater than or equal to $$0$$, that is,
$$\Delta = b^2-(y-a)(2y-6) \geq  0$$
$$y²-(a+3)y+3a-\dfrac{b^2}{8} \leq 0$$n
Since $$y$$ has minimum value 2 and maximum value 6
$$2\leq y\leq 6 $$ 
$$(y-2)(y-6)\leq 0 $$
Expand the inequality
$$y²-8y+12≤0$$n
Comparing (2) with (1) results in the system of equations in terms $$a,b $$
$$\begin{cases} a+3 &=8 \ 3a-\dfrac{b^2}{8}&=12 \end{cases}$$
Solving for $$a,b$$ yields
$$a = 5, b = \pm2\sqrt{6} $$

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