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by: aarellanes3

29

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1

# Chapter 2 15779

aarellanes3
UTEP
Precalculus
Victor Hugo Jimenez Nava

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Sections 2.1 -2.3
COURSE
Precalculus
PROF.
Victor Hugo Jimenez Nava
TYPE
Class Notes
PAGES
1
WORDS
KARMA
25 ?

## Popular in Applied Mathematics

This 1 page Class Notes was uploaded by aarellanes3 on Thursday October 1, 2015. The Class Notes belongs to 15779 at University of Texas at El Paso taught by Victor Hugo Jimenez Nava in Summer 2015. Since its upload, it has received 29 views. For similar materials see Precalculus in Applied Mathematics at University of Texas at El Paso.

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Date Created: 10/01/15
Chapter 2 1 Section 21 continued 393 To find the vertex you must use b2afb2a gt Major characteristic is that they are a parabola I When a is greater than 0 it opens upward I When a is less than 0 it opens downward I When its case 1 meaning upward then it has a lowest point I When its case 2 meaning downward then it has a highest point 393 Example gt Y2xquot24xl6 I Vertex422fb2a I X1 I Y10 I Vertex110 393 Now that the vertex has been solved lets solve for the intercepts gt Xbb4ac2a I a 2 I b4 I c 16 gt 444216 4 4 4 32 I 44128 4 124 gt We have simplified it all the way I Now we have two cases want to know why 0 Because of the that we had I Case 1 X4124 Case2 X4124 393 For the yintercept your solving for f0 remember yfx 115tandard form of a parabola st Basically if it39s a quadratic function the standard form will be gt fxaxhquot2k Vertex is h k 0 Treat h as x and k as y st Completing the square will come in handy when a function is in the form fxxquot2pxp2quot2

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