What is the range of (3x^2-12x+13)/(x^2-4x+4) ?

The range of (3x^2-12x+13)/(x^2-4x+4) is f(x)>3

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range of (3x^2-12x+13)/(x^2-4x+4)

range

\frac{3 x ^{2} - 12 x + 13}{x ^{2} - 4 x + 4}

f (x) > 3

Interval Notation

(3, \infty)

Solution steps

Rewrite as

\frac{3 x ^{2} - 12 x + 13}{x ^{2} - 4 x + 4} = y

Multiply both sides by

x^{2} - 4 x + 4

3 x^{2} - 12 x + 13 = y (x^{2} - 4 x + 4)

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

Discriminant

3 x^{2} - 12 x + 13 = y (x^{2} - 4 x + 4) : 4 y - 12

4 y - 12 \geq 0 : y \geq 3

Check if the range interval endpoints are included

y = 3 :

Excluded

Therefore the range is

f (x) > 3

y = - 125 x + 850

s^{2} + 2 s + 2

x^{3} - 40 x^{2} + 400 x

3 x - 1

f (x) = \frac{x - 4}{3 x - x ^{2}}

What is the range of (3x^2-12x+13)/(x^2-4x+4) ?
The range of (3x^2-12x+13)/(x^2-4x+4) is f(x)>3