How do you find the range of the function #y=f(x)=x/(x^2-5x+9)#?

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Answer 1 · 2015-04-22T19:15:33

The answer is : #-1/11<=y<=1#

Solution
#y=x/(x^2-5x+9)#

#=>yx^2-5yx+9y=x#

Send all terms to the left

#=> yx^2+(-1-5y)x+9y=0#

For real roots, #b^2-4ac>=0#

#=> (-1-5y)^2-4(y)(9y)>=0#

#=> 1+10y-11y^2>=0#

Now, all you've got to do is to factorize and find the range :)

#=>(-1-11y)(-1+y)>=0#

Or # (11y+1)(y-1)<=0#

Hence #-1/11<=y<=1#