6) - -2+4メ-5 1 --2 (x-2x)-5 = Coefcientofメー42 FE)--2(x²-2x+4)-5+4 H)= -2 (x-2)-1RIam getting this But the answeris f&) = -2(x-2)-3 why?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 31E
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Find the vertex form of each quadratic function by completing the square.

I attach two picture showing how I did the both problems. For #2 question I just want you to check if I did it right. And #5 I am getting a different answer can you show me step by step how you got the answer.

 

Thank You

 

ast 2 {)x16ス
Coefficient of x=16
- 64
%3D
FG-+16x+64)-64
f6)=(メ-8-GH
So, wertex form of fG) =x?+I6x islfG«)-(x-8)-64
Transcribed Image Text:ast 2 {)x16ス Coefficient of x=16 - 64 %3D FG-+16x+64)-64 f6)=(メ-8-GH So, wertex form of fG) =x?+I6x islfG«)-(x-8)-64
6 Fx) = -2x+4x-5
= Coefficientof *=42
%3D
PE)--2x²-2x+4)-5+4
=-2 (xー2)-1-エan geting this
But the answeris f&) =-2(x-2)-3 why?
Transcribed Image Text:6 Fx) = -2x+4x-5 = Coefficientof *=42 %3D PE)--2x²-2x+4)-5+4 =-2 (xー2)-1-エan geting this But the answeris f&) =-2(x-2)-3 why?
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