What is the range of (4x^2-16x+17)/(x^2-4x+4) ?

The range of (4x^2-16x+17)/(x^2-4x+4) is f(x)>4

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range of (4x^2-16x+17)/(x^2-4x+4)

range

\frac{4 x ^{2} - 16 x + 17}{x ^{2} - 4 x + 4}

f (x) > 4

Interval Notation

(4, \infty)

Solution steps

Rewrite as

\frac{4 x ^{2} - 16 x + 17}{x ^{2} - 4 x + 4} = y

Multiply both sides by

x^{2} - 4 x + 4

4 x^{2} - 16 x + 17 = y (x^{2} - 4 x + 4)

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

Discriminant

4 x^{2} - 16 x + 17 = y (x^{2} - 4 x + 4) : 4 y - 16

4 y - 16 \geq 0 : y \geq 4

Check if the range interval endpoints are included

y = 4 :

Excluded

Therefore the range is

f (x) > 4

x + 6 y = 2 y - 7

\frac{x - 3}{x ^{2} - x - 6}

\frac{3}{4} 2 x - 7 + 2

3 x - 2 y = - 6

x - y = 6

What is the range of (4x^2-16x+17)/(x^2-4x+4) ?
The range of (4x^2-16x+17)/(x^2-4x+4) is f(x)>4