math help! Consider the relation y= I2x-3I -1, where x is the independent variable.?

I need to graph the relation. Where do I start? I have the answer in the math workbook, but I want to learn how to do the work. Do I pick any values for x?

Top 4 Answers

A few days ago

John V

Favorite Answer

First, recognize that since I 2x – 3 I is always positive, y has a minimum at -1 when 2x-3 = 0, which is x = 3/2. For values of x > 3/2 or x < 3/2, y will be increasing from -1 linearly. Y = 0, for example, when 2x-3 = 1, which is x = 1 and x = 2. Similarly, y = 2 when x = 0 and x = 3. So you have the point (3/2, -1) as a minimum, and then (1,0), (2,0), and also (0,2), (2,3). Your graph will be v-shaped from the point (3/2,-1) with slope 2 going rightwards as the x-values increase, and slope -2 going leftwards as the x-values decrease, with a y -intercept of +2.

A few days ago

TurtleFromQuebec

Set up a piece of graph paper with x and y axes. then plug in some values for x, see what comes out for y, put those dots on the graph, and then see if you can connect the dots to make something that makes sense.

Certain points will be important since the function does something special at those points. One of the tricks of graphing (and functions in general) is to find these points. In your function an important point is x = 3/2 since this is where the absolute value is 0. So start from there and work left & right.

A few days ago

Sound_of_the_Bastard

the best way is to pick the values of x around 0, so -3 to 3 will be sensible since it will probably give the important points of the graph. Another way is to work out the X and Y intercepts of the function.

6 years ago

Anonymous

i dunno

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math help! Consider the relation y= I2x-3I -1, where x is the independent variable.?

math help! Consider the relation y= I2x-3I -1, where x is the independent variable.?

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