What is the range of y=(-3)/(12-x-x^2) ?

The range of y=(-3)/(12-x-x^2) is f(x) 0

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range of y=(-3)/(12-x-x^2)

range

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

f (x) \leq - \frac{12}{49} or f (x) > 0

Interval Notation

(- \infty, - \frac{12}{49}] \cup (0, \infty)

Solution steps

Rewrite as

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

Simplify

\frac{- 3}{12 - x - x ^{2}} : - \frac{3}{12 - x - x ^{2}}

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

Multiply both sides by

12 - x - x^{2}

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

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

Discriminant

- 3 = y (12 - x - x^{2}) : 49 y^{2} + 12 y

49 y^{2} + 12 y \geq 0 : y \leq - \frac{12}{49} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{12}{49} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{12}{49} or f (x) > 0

y = \frac{x}{x - 2}

f (x) = \frac{7}{4 - 2 x}

g (x) = \frac{2 x}{x ^{2} + 2 x - 24}

f (x) = 2 x^{2} + 8 x - 34

f (x) = \frac{1}{- x + 3}

What is the range of y=(-3)/(12-x-x^2) ?
The range of y=(-3)/(12-x-x^2) is f(x)<=-12/49 \lor f(x)>0