Question

Graph the inequality y<= -(2/3)|x-3|+4 Please show how

Answer 1

Answer:

Step-by-step explanation:

Definitions:

The absolute value equation in vertex form, given by: y = a|x – h| + k

where:

(h, k) = vertex

a  = determines whether the graph opens up or down. If a > 1, the graph opens up. If a < 1, the graph opens down.

h = determines how far left or right the graph is translated.

k = determines how far up or down the graph is translated.

The absolute value inequalities are graphed the same way as graphing the absolute value equations.  

Given the absolute value inequality, y ≤ - ⅔ |x - 3| + 4

where:

a = - ⅔

h = 3

k = 4

In reference to the definitions provided in this post, the vertex is (h, k). In the given inequality statement, the vertex occurs at point (3, 4).  Given the value of a = - ⅔, it means that the graph opens down. To find other points on the graph, we can solve for the intercepts.

Finding the Intercepts to Graph:

The y-intercept is the point on the graph where it crosses the y-axis. It is also the value of y when x = 0.  

To solve for the y-intercept, set x = 0:

y = - ⅔ |0 - 3| + 4

y = - ⅔ |- 3| + 4

y = - 2 + 4

y = 2  ⇒ y-intercept:  (0, 2).

The x-intercept is the point on the graph where it crosses the x-axis. It is also the value of x when y = 0.  

To solve for the x-intercept, set y = 0:

0 = - ⅔ |x - 3| + 4

0 - 4 = - ⅔ |x - 3| + 4 - 4

-4 = - ⅔ |x - 3|

-4 (3) =  (- ⅔ |x - 3| ) (3)

-12 = -2 |x - 3|

Divide both sides by -2:

$\frac{-12}{-2} = \frac{-2 |x - 3| }{-2}$

|x - 3| = 6

Apply absolute rule:

x - 3  = 6  or  x - 3 = - 6

x - 3 + 3 = 6 + 3    or  x - 3 + 3 = - 6 + 3

x = 9 or  x = -3

Therefore, the x-intercepts are:  (-3, 0) and (9, 0).

Graphing steps:

You now have the following points to plot on the graph:

vertex: (3, 4)

y-intercept:  (0, 2)

x-intercepts:  (-3, 0) and (9, 0).

You could easily connect these points with lines.  In graphing the boundary line, you will use a solid line due to the inequality symbol, "≤."  

The last step involves shading the appropriate half-plane region. In order to do this, choose a test point that is not on the boundary lines.  We can choose point, (0, 0). Substitute these coordinates into the absolute value inequality. If it provides a true statement, then you'll shade the region that contains that test point.

Test point: (0, 0)

y ≤ - ⅔ |x - 3| + 4

0 ≤ - ⅔ |0 - 3| + 4

0 ≤  $-\frac{2|0 - 3|}{3} + 4$

0 ≤  $-\frac{|- 6|}{3}$  + 4

0 ≤ - 2 + 4

0 ≤ 2 (True statement. Shade the region where (0, 0) is included).

Attached is a screenshot of the graphed absolute value inequality.  

Answer 2

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