Step 1Solution:-The given equation is y=|x+3|−1Step 2y=|x+3|1y=|(x+3)|−1⇒−x−3−1⇒−x−4;ifx+3&lt;0⇒xIt−3(x+3)−1=x+3−1=x+2;ifx+3≥0⇒x≥−3∣The graph of the function isa)vertex x=(−3,−1)b)Axis of symmetry x=−3c)zeros x=−2 and x=−4d)y=|x+3|−1 lies on 1st,2nd and 3rd quadrants and vertex lies on 3rd quadrante)range =(−∞,∞)f)y-intercepts =2f)slope for x≥3y=x+2compare withy=mx+cslope m=1for xIt3y=−x−4compare withy=m′xslopem′=−1g)Domain x=[−1,∞)

Solved: "y=|x+3|−1"

High School
Algebra
Pre-Algebra
Ratios, rates, proportions

Pizzadililehz

Answered question

2022-04-02

y = | x + 3 | - 1

Answer & Explanation

Ben Castillo

Beginner2022-04-03Added 13 answers

Step 1
Solution:-The given equation is

y = | x + 3 | - 1

Step 2

y = | x + 3 | 1

y = | (x + 3) | - 1 \Rightarrow - x - 3 - 1 \Rightarrow - x - 4; if x + 3 < 0 \Rightarrow x I t - 3

(x + 3) - 1 = x + 3 - 1 = x + 2; if x + 3 \geq 0 \Rightarrow x \geq - 3 ∣

The graph of the function is

a)vertex

x = (- 3, - 1)

b)Axis of symmetry

x = - 3

c)zeros

x = - 2

and

x = - 4

y = | x + 3 | - 1

lies on

1^{s t}, 2^{n d}

and

3^{r d}

quadrants and vertex lies on

3^{r d}

quadrant
e)range

= (- \infty, \infty)

f)y-intercepts

= 2

f)slope for

x \geq 3

y = x + 2

compare with

y = m x + c

slope

m = 1

for

x I t 3

y = - x - 4

compare with

y = m^{'} x

slope

m^{'} = - 1

g)Domain

x = [- 1, \infty)

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