SOLUTION: How do I graph this function v=x(4-2x) (6-2x) I am not sure where you receive the points from. Thanks, Christine

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Question 38397: How do I graph this function v=x(4-2x) (6-2x) I am not sure where you receive the points from. Thanks, Christine
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
How do I graph this function y = x(4 - 2x)(6 - 2x) I am not 
sure where you receive the points from.
Thanks, Christine


First find the three x-intercepts by substituting 0 for y

y = x(4 - 2x)(6 - 2x)

0 = x(4 - 2x)(6 - 2x)

Use the zero-factor principle, that is set each factor
equal to 0 and solve for x

The factors are x, (4 - 2x) and (6 - 2x)

Setting x = 0, just gives x = 0

Setting 4 - 2x = 0 gives x = 2

Setting 6 - 2x = 0 gives x = 3

So the three x-intercepts are (0, 0), (2, 0), and (3, 0)

Plot those.  Now we choose a value of x just less than x = 0, 
say x = -1, and substitute it for x in

y = x(4 - 2x)(6 - 2x)

y = -1(4 - 2(-1))(6 - 2(-1)) 

y = -1(4 + 2)(6 + 2)

y = -1(6)(8)

y = -48

So plot the point (-1, -48)

Now we choose a value of x between x = 0 and x = 2, 
say x = 1, and substitute it for x in

y = x(4 - 2x)(6 - 2x)

y = 1(4 - 2(1))(6 - 2(1)) 

y = 1(4 - 2)(6 - 2)

y = 1(2)(4)

y = 8

So plot the point (1, 8)

Now we choose a value of x between x = 2 and x = 3, 
say x = 2.5, and substitute it for x in

y = x(4 - 2x)(6 - 2x)

y = 2.5(4 - 2(2.5))(6 - 2(2.5)) 

y = 2.5(4 - 5)(6 - 5)

y = 2.5(-1)(1)

y = -2.5

So plot the point (2.5, -2.5)

Finally we choose a value of x just more than x = 3, 
say x = 4, and substitute it for x in

y = x(4 - 2x)(6 - 2x)

y = 4(4 - 2(4))(6 - 2(4)) 

y = 4(4 - 8)(6 - 8)

y = 4(-4)(-2)

y = 32

So plot the point (4, 32)

Now draw a smooth curve through those points moving from left to right:

Connect (-1,-48) to (0,0)
Connect (0,0) to (1,8)
Connect (1,8) to (2,0)
Connect (2,0) to (2.5,-2.5)
Connect (2.5,-2.5) to (3,0)
Connect (4,0) to (4,32)



Edwin McCravy
AnlytcPhil@aol.com