Question

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

Collected in the board: Fractions

Steven Zheng posted 7 hours ago

Answer

Rearrange the function

ax^2+bx+6 = y(x^2+2)

then,

(y-a)x²-bx+2y-6=0

which implies y\ne 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
(1)

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
(2)

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}


Steven Zheng posted 7 hours ago

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