Solving Quadratic Functions: A Quick Guide

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Homework Help Overview

The discussion revolves around understanding and manipulating functions, particularly in the context of quadratic and root functions. Participants are exploring the implications of transformations and the nature of the given equations.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the original poster's intent with the function y=-3f(2-x)-2 and whether it relates to quadratic functions. There are questions about the meaning of "root function" and the correct interpretation of transformations applied to the function. Some suggest isolating x in terms of y as a potential approach.

Discussion Status

The discussion is ongoing with multiple interpretations being explored. Some participants are providing guidance on how to approach the problem, while others express confusion about the original poster's statements and the lack of clarity in the problem setup.

Contextual Notes

There is uncertainty regarding the definitions and terminology used, such as "root function" and the relationship between the functions mentioned. Additionally, there are concerns about notation clarity in the context of square roots.

Nelo
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Homework Statement



y=-3f(2-x)-2 for the root function...

Homework Equations





The Attempt at a Solution



If I am given a function like, y=-3f(2-x)-2 for the root function...

Do i factor out the negetive on the x?

Making it.. y=-3f(-(x-2)-2 . making a reflection on both the x/y axis's ? Or is that wrong?
 
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Simple yes/no answer? :)
 
I don't want to be rude, but I have no idea what you're trying to do. Is y = f(x)? Is your equation:

f(x) = 3f(2-x) - 2?

What are you trying to do with this?
 
"Yes, No" to what question? You titled this "Quick Quadratics Question", but there is no quadratic in the problem. You say you are given the "root" function -3(2- x)- 2. What do you mean by "root function"? and , as gb7nash aked what are you trying to do with it?
 
Perhaps the thing to do is to solve for x in terms of y? I don't fully understand this either. You could isolate x on the left and that might be what is being asked? Not sure? will look forward to the correct solution.
 
::: (sqrt)2(x-4) +1

Is this a horizontal or vertical stretch/compression?

I said it was a vertical, since its outside the bracket.
 
Nelo said:
::: (sqrt)2(x-4) +1

Is this a horizontal or vertical stretch/compression?

I said it was a vertical, since its outside the bracket.
Please stop using your (sqrt) notation, with parenthes around "sqrt". Put the parentheses around the expression whose square root you're taking. With your notation it's impossible to tell whether the +1 is inside the radical or outside.

Is this your function?
y = sqrt(2(x - 4) + 1)
 

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