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o-na [289]
2 years ago
7

What is the graph of the absolute value inequality?|x + 5| ≥ y – 2

Mathematics
2 answers:
andrew11 [14]2 years ago
7 0
Best way to describe the answer is that the point will touch -6 and the left top corner will be shaded in
Lilit [14]2 years ago
6 0

Step-by-step explanation:

|x + 5| ≥ y – 2

First we replace the inequality sign by = sign

|x + 5| = y – 2

solve for y and then graph it

add 2 on both sides

y = |x + 5|+ 2

Equation is in the form of y=a|x-h|+k

where (h,k) is the vertex

h = -5  and k = 2

So vertex is (-5,2)

Now make a table

x            y = |x + 5|+ 2

-10         y = |-10+5| + 2= 7

-5            y = |-5+5| + 2= 2

0              y = |0+5| + 2= 7

Now plot all the points we got the table and graph it

Now we do shading

Lets pick any point inside the V shaped graph and check with our inequality

LEts pick (-5,5)

|x+5| ≥ y – 2

plug in -5 for x  and 5 for y

|-5+5| ≥ 5– 2

0 > = 3, its false

So we cannot shade the part containing (-5,5)

we shade the bottom part on graph

the graph is attached below


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<h3>Answer: C) -1/2</h3>

==================================

Work Shown:

We will use the general vertex form y = a(x-h)^2+k

(h,k) is the vertex, and the 'a' value stretches or compresses the graph vertically

(h,k) = (-2,18) since the highest point is at (-2,18)

Use either root -8 or 4 to plug into the equation as well. I'll use -8

The root -8 means the point (-8,0) is an x intercept.

----------------

Plug the values mentioned into the equation below. Then solve for 'a'.

y = a(x-h)^2+k

y = a(x-(-2))^2+18 ... plug in (h,k) = (-2,18)

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0 = a(-8+2)^2+18 ... plug in (x,y) = (-8,0)

0 = a(-6)^2+18

0 = a*36+18

0 = 36a+18

36a+18 = 0

36a = -18 ... subtract 18 from both sides

a = -18/36 .... divide both sides by 36

a = (-1*18)/(2*18)

<h3>a = -1/2</h3>

---------------

Alternatively, you could use (x,y) = (4,0), instead of (x,y) = (-8,0), and you'll get the same 'a' value as well.  

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Answer:

see explanation

Step-by-step explanation:

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\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

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