FROM THE GIVEN ABSOLUTE VALUE EQUATIONS FIND VERTEX

Compare the absolute value function with the parent function and find

(i) vertex  (ii)  a  (iii)  opens up or down

Example 1 :

y  =  6|x-7|

Solution :

By comparing the given absolute value function with parent function

y  =  a|x-h| + k

y  =  6|x-7| + 0

We get,

Vertex (h, k)  ==>  (7, 0)

The absolute value function opens at (7, 0).

a  =  6/1

For every 6 units rise, 1 unit run. Since it is positive, it opens up.

Example 2 :

y   =  -|x-8| + 1

Solution :

Vertex (h, k)  ==>  (8, 1)

The absolute value function opens at (8, 1).

a  =  -1/1

For every 1 unit rise, 1 unit run. Since it is negative, it opens down.

Example 3 :

y   =  (1/3)|x-3| + 4

Solution :

Vertex (h, k)  ==>  (3, 4)

The absolute value function opens at (3, 4).

a  =  1/3

For every 1 unit rise, 3 units run. Since it is positive, it opens up.

Example 4 :

y   =  |x| - (5/2)

Solution :

Vertex (h, k)  ==>  (0, -5/2)

The absolute value function opens at (0, -5/2).

a  =  1/1

For every 1 unit rise, 1 unit run. Since it is positive, it opens up.

Example 5 :

y   =  |x| + 9

Solution :

Vertex (h, k)  ==>  (0, 9)

The absolute value function opens at (0, 9).

a  =  1/1

For every 1 unit rise, 1 unit run. Since it is positive, it opens up.

Example 6 :

y   =  -|x+2| + 11

Solution :

Vertex (h, k)  ==>  (-2, 11)

The absolute value function opens at (-2, 11).

a  =  -1/1

For every 1 unit rise, 1 unit run. Since it is negative, it opens down.

Example 7 :

y   =  -2|x+9| + 3

Solution :

Vertex (h, k)  ==>  (-9, 3)

The absolute value function opens at (-9, 3).

a  =  -2/1

For every 2 units rise, 1 unit run. Since it is negative, it opens down.

Example 8 :

y   =  (-1/2)|x+6|

Solution :

Vertex (h, k)  ==>  (-6, 0)

The absolute value function opens at (-6, 0).

a  =  -1/2

For every 1 unit rise, 2 units run. Since it is negative, it opens down.

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