anonymous
  • anonymous
Graph the equation written:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[f(x)=\frac{ 1+|3-x| if x is \neq 3 }{ 2 if x =3 }\]
austinL
  • austinL
What? \(f(x)=\dfrac{1+|3-x|}{2},~\text{if x}\ne3,~ \text{if x}=3\) Is this the question?
anonymous
  • anonymous
Yes but the x unequal to 3 goes for the top and the x=3 is for the denominator

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austinL
  • austinL
I have never seen an equation in that form where the location of.... just.... um. @nincompoop Would you mind looking at this?
anonymous
  • anonymous
Alright, Well see what this person says. Lol
anonymous
  • anonymous
Its a part of limits with calculus
austinL
  • austinL
I guess nin is busy.
anonymous
  • anonymous
Its alright. Wanna help with another one?
anonymous
  • anonymous
the function doesnt have a denominator right??
anonymous
  • anonymous
its not a function. its for a limit to be graphed. Look at the very first one @hantenks
anonymous
  • anonymous
The fraction was to keep them separated.
anonymous
  • anonymous
\[F(x)= 1+|3-x|; if x \neq 3\]
anonymous
  • anonymous
\[Then 2 if x=3\]
austinL
  • austinL
THAT is what confused me.
anonymous
  • anonymous
Ohhhh, I'm sorry. I didn't know.
anonymous
  • anonymous
But its asking for it to be graphed.
anonymous
  • anonymous
ohkzz.. its simple if you can graph the 1st part of f(x) at x not equal to 0.. for it you need to know how to draw |x| graph m posting it..
DebbieG
  • DebbieG
So graph y=1+|3-x| which is just a transformation of an abs value function. But, at x=3, you need an "open circle" for the value you get for y, and then a closed circle at the point (3,2).
anonymous
  • anonymous
So what exactly would I want to do first. |dw:1378749273650:dw|
DebbieG
  • DebbieG
Sorry, I don't understand that^^ at all. Do you know how to graph y=1+|3-x| ?
anonymous
  • anonymous
No thats why I asked. I wasn't forsure how to graph the |3-x| part
DebbieG
  • DebbieG
OK, then lets start there. Do you know how to graph y=|x|?
anonymous
  • anonymous
|dw:1378750942088:dw|
DebbieG
  • DebbieG
y=|-x| is a reflection over the y-axis of y=|x|... so doesn't change a thing. y=|3-x| is a horizontal shift (can you tell me which direction? y=1+|3-x| is a vertical shift (can you tell me which direction?)
anonymous
  • anonymous
Ok then we would want to try that with the 3-x then right?
DebbieG
  • DebbieG
Yes, start with going from that ^^ (he did the first part, y=|x|, for you) to y=|3-x| It's a horizontal shift.
anonymous
  • anonymous
|dw:1378749474081:dw|
anonymous
  • anonymous
Ok, then we need to add the shift right?
DebbieG
  • DebbieG
Right, one step at a time. yes, good - shift right.
anonymous
  • anonymous
Shift right and then go over 3 right
anonymous
  • anonymous
like graphing a slope
DebbieG
  • DebbieG
Then, as you said, shift up 1 unit. That gives you y=1+|3-x| Now you just "erase" the point on that graph at x=3.... you can indicate that with a small "open circle" at the point on the graph for x=3 (3,1). Then put in the point that IS part of the graph for x=3, which is (3,2). Just plot that lonely little point.
DebbieG
  • DebbieG
I don't know what you mean by "Shift right and then go over 3 right" You shift right, 3 units. That's the horizontal shift. that's ONE transformation. Then you shift up, 1 unit, that's the vertical shift.
anonymous
  • anonymous
So there is no lines? Sorry I need to leave but let me know
anonymous
  • anonymous
It would be the two points techniqually
DebbieG
  • DebbieG
|dw:1378749759389:dw|
anonymous
  • anonymous
Awesome
DebbieG
  • DebbieG
Well, I'm not sure what you mean.... maybe you haven't done graphing functions with transformations yet? If so, then we need to slightly change the approach. You can just graph the absolute value function "as if" it was a piecewise function, itself.... so you would have y=1+|3-x| graphed as: y=-x+4 when x<3 y=x-2 when x>3 So you can view them as 2 different lines. That is an alternative approach. And again, open dot where they meet at (3,1), and a closed dot for the point at (3,2)

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