Math Help - Finding range of complex quadratic

1. Finding range of complex quadratic

Hi there,

I need to find the range algebraically of the following equation:

(x2-6x+5)/(x2+2x+1)

I know the answer is everything greater than -1/3.

How do I do this?

2. Re: Finding range of complex quadratic

Are you allowed to use differential calculus?

3. Re: Finding range of complex quadratic

Hi there,

I need to find the range algebraically of the following equation:

(x2-6x+5)/(x2+2x+1)

I know the answer is everything greater than -1/3.

Did you notice that $x^2+2x+1=(x+1)^2~\&~x^2-6x+5=(x-1)(x-5)~?$

Did you draw a graph?

4. Re: Finding range of complex quadratic

Yes, I know you can factorise to give (x-1)(x-5)/(x+1)(x+1), and I deduced the lower limit of y being -1/3 from plotting the graph.

However is there a way of getting -1/3 algebraically?

5. Re: Finding range of complex quadratic

what is the right hand side of the equation, unless you give that it is not an equation. If we consider it as a function i.e., f(x) = (x2-6x+5)/(x2+2x+1)
Then we get f(x) = [ ( x-5) ( x-1) ]/(x+1)^2. Notice the function is not defined for x = -1, thus domain is R-{1} and range is R

6. Re: Finding range of complex quadratic

Originally Posted by ibdutt
what is the right hand side of the equation, unless you give that it is not an equation. If we consider it as a function i.e., f(x) = (x2-6x+5)/(x2+2x+1)
Then we get f(x) = [ ( x-5) ( x-1) ]/(x+1)^2. Notice the function is not defined for x = -1, thus domain is R-{1} and range is R
Actually when graphing the function, Range is everything greater than -1/3. Your domain is correct in everything except -1 where the asymptote is.

7. Re: Finding range of complex quadratic

If anyone is interested I've figured out how to do this after research.

(x2-6x+5)/(x2+2x+1)

You first make a new quadratic equation using the coefficient of the denominator times y, minus the coefficient of the numberator.

So (y-1)x^2+(2y+6)x+(y-5)

You then use these terms to find the discriminant.

Sqrt (2y+6)^2-4(y-1)(y-5)

You do the expanding and are left with 48y+16, which then becomes y = 1/3. You take the negative of this and that becomes the lower limit and hence the minimum value for y.

If you get two solutions for y, the range is between those two values.

8. Re: Finding range of complex quadratic

It looks like what you have done is:

$y=\frac{x^2-6x+5}{x^2+2x+1}$

$y(x^2+2x+1)=x^2-6x+5$

$(y-1)x^2+(2y+6)x+(y-5)=0$

Now, find the axis of symmetry:

$x=-\frac{2y+6}{2(y-1)}=-\frac{y+2}{y-1}$ and substitute this value for $x$ to get:

$\frac{(y+3)^2}{y-1}-\frac{2(y+3)^2}{y-1}+y-5=0$

$(y-5)(y-1)=(y+3)^2$

$y^2-6y+5=y^2+6y+9$

$12y=-4$

$y=-\frac{1}{3}$