What is the range of (x^2-2x-63)/(x+9) ?

The range of (x^2-2x-63)/(x+9) is f(x) =-8

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range of (x^2-2x-63)/(x+9)

range

\frac{x ^{2} - 2 x - 63}{x + 9}

f (x) \leq - 32 or f (x) \geq - 8

Interval Notation

(- \infty, - 32] \cup [- 8, \infty)

Solution steps

Rewrite as

\frac{x ^{2} - 2 x - 63}{x + 9} = y

Multiply both sides by

x + 9

x^{2} - 2 x - 63 = y (x + 9)

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

Discriminant

x^{2} - 2 x - 63 = y (x + 9) : y^{2} + 40 y + 256

y^{2} + 40 y + 256 \geq 0 : y \leq - 32 or y \geq - 8

Check if the range interval endpoints are included

y = - 32 :

Included

y = - 8 :

Included

Therefore the range is

f (x) \leq - 32 or f (x) \geq - 8

f (x) = x^{3} - 27 x

g (x) = - 4 x + 1

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

(y) = - 3 x + 7

f (x) = 3 x^{2} - 6 x

What is the range of (x^2-2x-63)/(x+9) ?
The range of (x^2-2x-63)/(x+9) is f(x)<=-32\lor f(x)>=-8