What is the range of (x^2-6x+12)/(x-4) ?

The range of (x^2-6x+12)/(x-4) is f(x) = 6

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range of (x^2-6x+12)/(x-4)

range

\frac{x ^{2} - 6 x + 12}{x - 4}

f (x) \leq - 2 or f (x) \geq 6

Interval Notation

(- \infty, - 2] \cup [6, \infty)

Solution steps

Rewrite as

\frac{x ^{2} - 6 x + 12}{x - 4} = y

Multiply both sides by

x - 4

x^{2} - 6 x + 12 = y (x - 4)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x^{2} - 6 x + 12 = y (x - 4) : y^{2} - 4 y - 12

y^{2} - 4 y - 12 \geq 0 : y \leq - 2 or y \geq 6

Check if the range interval endpoints are included

y = - 2 :

Included

y = 6 :

Included

Therefore the range is

f (x) \leq - 2 or f (x) \geq 6

f (x) = x - 3

y = \frac{x + 8}{x}

f (x) = \frac{2}{5} x - 4

f (x) = x^{3} + 8

f (x) = \frac{6}{x} + 9

What is the range of (x^2-6x+12)/(x-4) ?
The range of (x^2-6x+12)/(x-4) is f(x)<=-2\lor f(x)>= 6