What is the y=(|x+1|-3)/(1+|x-3|),-2<= x<4 ?

The y=(|x+1|-3)/(1+|x-3|),-2<= x<4 is

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y=(|x+1|-3)/(1+|x-3|),-2<= x<4

y = \frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4

Domain : - 2 \leq x < 4

Range : - \frac{3}{5} \leq f (x) \leq 1

X Intercepts : (2, 0), Y Intercepts : (0, - \frac{1}{2})

Asymptotes : Horizontal None

Extreme Points : Maximum (- 2, - \frac{1}{3}), Minimum (- 1, - \frac{3}{5})

Interval Notation

Range : [- \frac{3}{5}, 1]

Solution steps

Domain of

\frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4 : - 2 \leq x < 4

Range of

\frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4 : - \frac{3}{5} \leq f (x) \leq 1

Axis interception points of

\frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4 :

X Intercepts

: (2, 0),

Y Intercepts

: (0, - \frac{1}{2})

Asymptotes of

\frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4 :

Horizontal

:

None

Extreme Points of

\frac{∣ x + 1 ∣ - 3}{1 + ∣ x - 3 ∣}, - 2 \leq x < 4 :

Maximum

(- 2, - \frac{1}{3}),

Minimum

(- 1, - \frac{3}{5})

f (x) = 3^{(3 x + 1)}

f (x) = - 4 x^{2} - 9

f (x) = arcsin (3 x - 4 x^{3})

f (x) = 5 x^{2} + 7 x - 6

f (x) = \frac{x}{9} + 1

What is the y=(|x+1|-3)/(1+|x-3|),-2<= x<4 ?
The y=(|x+1|-3)/(1+|x-3|),-2<= x<4 is