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

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MarkFL

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Feb 24, 2012
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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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MarkFL

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Feb 24, 2012
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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.