Melody

2+((x^2)/(1-x))

$$2+\frac{x^2}{1-x}$$

To work out the domain, the idea is to work out what x can be.
It is usually easier to work out what x can't be though.

So with this one.
you cannot divide by zero so 

$$\\1-x\ne 0\rightarrow\;\;x\ne 1$$

Domain: All Real x where x is not equal to 1
or

Domain: All Real x except x=1

14x16+7.  Thats it

Hi Sabi

It looks good to me :)

Why didn't you see his answer?

The fact that you had an answer would have been clearly displayed in your watch list.  :/

Sabi92, if you do not understand Geno's answer, you need to post questions about it on the original thread. :)

This is a very strangely worded question - that is probably why no one has answered it already. :/

The point (1,-2) is on the graph of f(x). Describe the following transformations on f(x), and determine the resulting point.

g(x)=2f(x)+3          We know that f(1)=-2

so g(1)=2*-2+3 = -4+3 = -1

so every output value from the f function is doubled and then 3 is added to get the output for the g function.

g(x)=f(x+1)−3

g(0)=f(0+1)-3 =f(1)-3 = -2-3=-5

I don't know what the question wants - it is weird ://      Sorry

g(x)=−f(2x)

g(x)=−f(−x−1)+3

Thanks Alan,

OR

You could solve it by completing the square method

$$\\x^2-4xm+(2m-\frac{1}{4})=0\\\\
x^2-4xm\;=\;-(2m-\frac{1}{4})\\\\
x^2-4xm+(2m)^2\;=\;-2m+\frac{1}{4}+(2m)^2\\\\
(x-2m)^2\;=\;-2m+\frac{1}{4}+4m^2\\\\
(x-2m)^2\;=\;4m^2-2m+\frac{1}{4}\\\\
(x-2m)^2\;=\;4(m^2-\frac{m}{2}+\frac{1}{16})\\\\
(x-2m)^2\;=\;4(m-\frac{1}{4})^2\\\\
x-2m\;=\;\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m+ 2m-\frac{1}{2}\qquad or \qquad x\;=\;2m- 2m+\frac{1}{2}\\\\
x\;=\;4m-\frac{1}{2}\qquad or \qquad x\;=\;\frac{1}{2}\\\\$$

You would have to give a specific example.    

Zero

it will be zero

@@ End of Day Wrap  Sun 12/7/15   Sydney, Australia Time   8:25pm  ♪ ♫

Hi everyone,

It was another quiet day but there were still some good questions and our answerers were keen to help :)

Today answer credits go to TitaniumRome, CPhill, Geno3141, Radix and Alan.  Thank you :)