PRoblems with precalc functions

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Homework Help Overview

The discussion revolves around understanding the behavior of a quadratic function, specifically identifying intervals of increase and decrease, as well as determining the domain and range. The subject area includes precalculus functions and their properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find where the function increases and decreases after calculating the vertex. Some participants provide different vertex calculations and discuss the implications of the leading coefficient on the function's behavior.

Discussion Status

Participants are exploring different interpretations of the vertex and its significance in determining the function's increasing and decreasing intervals. There is some productive dialogue regarding the correct application of formulas, but no consensus has been reached on the original poster's questions.

Contextual Notes

One participant notes the importance of not mixing up the leading coefficient, which affects the function's characteristics. Additionally, a separate question about solving a trigonometric equation appears, but it is suggested that it should be addressed in a new thread.

crazy_shoes
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I'm trying to find where this function increases and decreases:

f(x) = = - x^2 - 6x - 8


I can get the vertex (h, k), which works out [by my math] to be (3, -17).

I'm just not sure where to go from here to get information like:
point of increase and decrease
domain and range

Thanks for any help!
 
Last edited:
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For the vertex, I get (-3,1), using the formula for h=-b/(2a). The vertex is where it switches from increasing to decreasing or from decreasing to increasing. The leading coefficient tells you which of these two cases it actually is...
 
Aaah... I see what I did wrong, I simply forgot that a=-1, I was thinking a=1.

When I set a to -1 instead of 1 it all works out! Thanks!
 
hi,
i need help finding 2 solutions to an equation. cos(theta) = 1/(sq.root of 2)
thanks !
 
Ramzi said:
hi,
i need help finding 2 solutions to an equation. cos(theta) = 1/(sq.root of 2)
thanks !

It's considered bad form to "hijack" an existing thread with a new question.

You should start a new thread.
 
im sorry ! its my first time using forums and all that stuff. my bad, but can anyone help me ?
 

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