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graph the following function using transformations be sure to graph all stages on one...

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stargsd5 | Student, College Freshman | Valedictorian

Posted August 14, 2012 at 6:53 PM via web

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graph the following function using transformations be sure to graph all stages on one graph. state the domain and the range

y=(1/4)(x+4)^2

show each stage , not just the final one

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted August 14, 2012 at 8:08 PM (Answer #1)

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The graph of `y=x^2`  is

 

`y=(x+4)^2` represents taking `y = x^2` and shifting it 4 units to the left:

  

`y=1/4 (x+4)^2` represents taking `y=(x+4)^2` and "squishing" the graph vertically by a factor of 4.  That is, where y used to be 4, now it is 1, where y used to be 8, now it is 2, where y used to be 1, now it is .25

  

 

Looking at those last two together on the same graph:

The squished version is the blue parabola, the unsquished one is the red parabola.  Notice at -3, the red one has height 1, the blue has a height of .25.  At -2, the red one has height 4, the blue has a height of 1.  At (approximately) -1.2, the red one has height 8, the blue one has height 2

 

 

The domain is all real numbers.  There is no x value that you can't plug into `y=1/4 (x+4)^2`

You can take any number and add 4 to it.  You can take any number and square it.  You can take any number and multiply by 1/4.  So the domain is all real numbers.

The range is all possible y values.  Looking at the graph, you can see that you can get all the positive values, but you can't get any negative ones.  There is nowhere on the graph where the height of the parabola is -3, for example.  

Another way to see this, is that x+4 can be any number (positive or negative).  But then you square it, and `(x+4)^2` must be either 0 or positive.  It can't be negative, because it is a square.  A positive number times 1/4 is still positive.  So the range is: `y>=0`

 

 

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