Graphing Quadratic Functions: Domain, Range, and Factoring Help

[Nelo]

Homework Statement

Given f(x) = x^2 Sketch the graph of each of the following, state the domain and range.



- 1/2f(2x+6) -2

Homework Equations

y=x^2

The Attempt at a Solution

I Simply have a question. The question is, that this 2x can be factored out from the brackets correct?

Factoring that 2 will multiply it with the 1/2 and make the vertical stretch a factor of 1 , esentially, a factor of 0. No?

Or , would it create -1/2f(2(x+3) -2 , where the horizontal still exists as 2 (1/2) and the vertical still exists as 1/2 ?

Which one is it?

 

[LCKurtz]

Homework Statement

Given f(x) = x^2 Sketch the graph of each of the following, state the domain and range.

- 1/2f(2x+6) -2

Homework Equations

y=x^2

The Attempt at a Solution

I Simply have a question. The question is, that this 2x can be factored out from the brackets correct?

Factoring that 2 will multiply it with the 1/2 and make the vertical stretch a factor of 1 , esentially, a factor of 0. No?

Or , would it create -1/2f(2(x+3) -2 , where the horizontal still exists as 2 (1/2) and the vertical still exists as 1/2 ?

Which one is it?

You need to use parentheses when writing your formulas so we can understand what your formula is. What all is in the denominator? And after the / your parentheses are unbalanced. Anyway, whatever you actually meant, no, you can not factor the two across the f, if that is what you are asking.

 

[eumyang]

I think he/she's asking if you can factor out the 2 in 2x + 6, and the answer is yes, you can, as long as you don't bring it outside the f.

I'm guessing that he/she meant to write this:
f(x) = -(1/2)f(2x+6) - 2 = -(1/2)f(2(x+3)) - 2
or this:
$f(x) = -\frac{1}{2}f(2x +6) - 2 = -\frac{1}{2}f(2(x +3)) - 2$

OP: Maybe you should learn LaTeX.