Functions

R

rachelmaddie

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I need my work checked please and I’m having trouble with graphing the function.
6C3211D0-12D3-4C7E-B3FB-D4355F2DCFD4.png
The given function is y = ((x + 4)-2))

Domain of the given function is: (x + 4) >_ 0
(Since the square root of numbers less than 0 is not possible)
Domain: x _> (-4)
Range: x_> -2

Domain: [-4, ∞)
Range: [-2, ∞)
 
rachelmaddie said:
I need my work checked please and I’m having trouble with graphing the function.
View attachment 20363
The given function is y = ((x + 4)-2))......................... incorrect

Domain of the given function is: (x + 4) >_ 0
(Since the square root of numbers less than 0 is not possible)
Domain: x _> (-4)
Range: x_> -2

Domain: [-4, ∞)
Range: [-2, ∞)
Click to expand...
 
I had trouble with that because of the notation. I didn’t know how to type it out with the brackets..
 
So, where are you having trouble graphing? Do you have an attempt to show?
 
Please somehow show us WHAT trouble you are having, so we don't have to give you an entire course on graphing.

Either show us an attempt at some graph, so we can see what is going wrong, or tell us specifically what step you find confusing.
 
Dr.Peterson said:
Please somehow show us WHAT trouble you are having, so we don't have to give you an entire course on graphing.

Either show us an attempt at some graph, so we can see what is going wrong, or tell us specifically what step you find confusing.
Click to expand...
Is this correct?F3329DF9-7CCF-4BE8-9561-03A13CA69001.jpeg
 
Yes. You shifted the graph of [MATH]y = \sqrt{x}[/MATH] left 4 units and down 2, which is correct. And you can check: Does the point (0, 0) satisfy the equation? How about (5, 1)?

Of course, this looks as if you just put the equation into Desmos ...
 
Dr.Peterson said:
Yes. You shifted the graph of [MATH]y = \sqrt{x}[/MATH] left 4 units and down 2, which is correct. And you can check: Does the point (0, 0) satisfy the equation? How about (5, 1)?

Of course, this looks as if you just put the equation into Desmos ...
Click to expand...
I’m drawing it myself. Is this good?301CF5D6-8C46-40AF-9F9E-1B72BA7F7E4C.jpeg
 
The important thing is that you got the starting point correctly (indicating the right translations), which is the main point of the exercise; but the shape looks too much like a line, so I might take a little off. The square root function actually starts of vertically before immediately turning downward.

To improve it, I would start by plotting a few familiar points of the original square root function ([MATH]\sqrt{1}[/MATH], [MATH]\sqrt{4}, ...[/MATH]) and then shifting those to plot points on the required graph.
 
Dr.Peterson said:
The important thing is that you got the starting point correctly (indicating the right translations), which is the main point of the exercise; but the shape looks too much like a line, so I might take a little off. The square root function actually starts of vertically before immediately turning downward.

To improve it, I would start by plotting a few familiar points of the original square root function ([MATH]\sqrt{1}[/MATH], [MATH]\sqrt{4}, ...[/MATH]) and then shifting those to plot points on the required graph.
Click to expand...
Can you help me with the drawing? I was having trouble trying to digitally draw the line properly.
 
Well, I can't hold your hand while you do it, so I'm not sure how I can help. Can you draw a curve through (0,0), (1,1), (4,2), and (9,3)?
 
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