SOLUTION: A curve has equation y = (x^2-5x+7)/(x-3). Find the range of y for real x. Hence, or otherwise, find the coordinates of the turning points.

Question 1075131: A curve has equation y = (x^2-5x+7)/(x-3).
Find the range of y for real x. Hence, or otherwise, find the coordinates of the turning points.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
y = (x^2-5x+7)/(x-3)
If x=0, y=-7/3
if x is infinite positive, y is infinite positive
If x is infinite negative, y is infinite negative
Can y=0?
No, because the discriminant, when you set the expression equal to 0 and multiply by x-3, is negative.
The derivative is (x^2-6x+8)/(x-3)^2. If you set that equal to 0, x equals 2 and 4.
As x approaches 3 from the positive side, y approaches + infinity.
As x approaches 3 from the negative side, y approaches - infinity.
The turning points are at x=2 and x=4, where y= -1 and and 3 respectively (2,-1) and (4, 3)
The range is (-oo,-1] and [3,oo)

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