MHB John's Question: Finding Range of Rational Function

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To find the range of the rational function (x^2-6x+9)/(2x-8), the process involves setting the equation in standard quadratic form and applying the discriminant condition for real roots. By rearranging the equation, it becomes y(2x-8) = x^2 - 6x + 9, leading to a quadratic in terms of y. The discriminant must be non-negative, resulting in the inequality y(y-2) ≥ 0, indicating that y is either less than or equal to 0 or greater than or equal to 2. Therefore, the range of the function is (-∞, 0] ∪ [2, ∞).
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Here is the question:

Find the range of (x^2-6x+9)/(2x-8).?

Hello. I was given this in an assignment recently. When my teacher handed out the answer sheet after he had marked the assignments. I was puzzled to find that he had found the range using the discriminant of the quadratic formula. I will denote Lambda as Y since i don't have a lambda button. Here is the working:

Y=(x^2-6x+9)/(2x-8)
Y(2x-8)=(x^2-6x+9)
X^2+(-2Y-6)x+9+8Y=0

(-2Y-6)^2-4(9+8Y) ≥0
and then through working eventually arrived at:
Y≤ 0 or Y ≥ 2.
I really don't get how he used this to find the range or how he got ≤ and ≥ in the final result. Please help I have a test coming up soon and this is the only thing I don't understand.

I have posted a link there to this topic so the OP can see my work.
 
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Hello John,

We are given to find the range of:

$$y=\frac{x^2-6x+9}{2x-8}$$

If we multiply through by $2x-8$, we have:

$$y(2x-8)=x^2-6x+9$$

Arranging in standard quadratic form, we obtain:

$$x^2-2(3+y)+(9+8y)=0$$

Now, in order for this equation to have real roots, we require the discriminant to be non-negative, hence:

$$(-2(3+y))^2-4(1)(9+8y)\ge0$$

Expanding, distributing and collecting like terms, we have:

$$4(3+y)^2-4(9+8y)\ge0$$

$$\left(9+6y+y^2 \right)-(9+8y)\ge0$$

$$9+6y+y^2-9-8y)\ge0$$

$$y^2-2y=y(y-2)\ge0$$

Since this is an upward opening parabola with roots at $y=0,2$, we know it is negative on $(0,2)$, and so it is non-negative on:

$$(-\infty,0]\,\cup\,[2,\infty)$$

And thus, this is the range of the original function.
 
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