# John's question at Yahoo! Answers regarding finding the range of a rational function

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
Hello John,

We are given to find the range of:

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

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

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

Arranging in standard quadratic form, we obtain:

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

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

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

Expanding, distributing and collecting like terms, we have:

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

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

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

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

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:

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

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