Graph the equation written:

- anonymous

Graph the equation written:

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- anonymous

\[f(x)=\frac{ 1+|3-x| if x is \neq 3 }{ 2 if x =3 }\]

- austinL

What?
\(f(x)=\dfrac{1+|3-x|}{2},~\text{if x}\ne3,~ \text{if x}=3\)
Is this the question?

- anonymous

Yes but the x unequal to 3 goes for the top and the x=3 is for the denominator

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## More answers

- austinL

I have never seen an equation in that form where the location of.... just.... um.
@nincompoop
Would you mind looking at this?

- anonymous

Alright, Well see what this person says. Lol

- anonymous

Its a part of limits with calculus

- austinL

I guess nin is busy.

- anonymous

Its alright. Wanna help with another one?

- anonymous

the function doesnt have a denominator right??

- anonymous

its not a function. its for a limit to be graphed. Look at the very first one @hantenks

- anonymous

The fraction was to keep them separated.

- anonymous

\[F(x)= 1+|3-x|; if x \neq 3\]

- anonymous

\[Then 2 if x=3\]

- austinL

THAT is what confused me.

- anonymous

Ohhhh, I'm sorry. I didn't know.

- anonymous

But its asking for it to be graphed.

- 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

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

So what exactly would I want to do first. |dw:1378749273650:dw|

- DebbieG

Sorry, I don't understand that^^ at all.
Do you know how to graph y=1+|3-x| ?

- anonymous

No thats why I asked. I wasn't forsure how to graph the |3-x| part

- DebbieG

OK, then lets start there.
Do you know how to graph y=|x|?

- anonymous

|dw:1378750942088:dw|

- 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

Ok then we would want to try that with the 3-x then right?

- 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

|dw:1378749474081:dw|

- anonymous

Ok, then we need to add the shift right?

- DebbieG

Right, one step at a time.
yes, good - shift right.

- anonymous

Shift right and then go over 3 right

- anonymous

like graphing a slope

- 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

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

So there is no lines? Sorry I need to leave but let me know

- anonymous

It would be the two points techniqually

- DebbieG

|dw:1378749759389:dw|

- anonymous

Awesome

- 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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