+0  
 
+1
54
7
avatar+461 

The quadratic \(ax^2 + bx + c \) can be expressed in the form \(2(x - 4)^2 + 8 \). When the quadratic \(3ax^2 + 3bx + 3c \) is expressed in the form \(n(x - h)^2 + k \), what is \(h \)?

 

I think its 3 but idk

 Oct 25, 2020
edited by HelpBot  Oct 25, 2020
 #2
avatar+28021 
+2

2 (x-4)^2 +8      expand to find  a    b   c

2x^2 -16x + 40           a = 2    b = -16   c = 40

 

then    3a = 6    3b = -48    3c = 120

 

6x^2 -48x +  120

6 (x-4)^2  +24                 h = -4

 

Which makes sense, because all you did was multiply the original expression by 3

3 * (2(x-4)^2 + 8) =    6(x-4)^2+24     h = -4

 Oct 25, 2020
 #6
avatar+28021 
0

Sorry....  my mistake      all of the math is correct  EXCEPT h = +4      NOT - 4  

Thanx Helpbot for pointing that out to me !

ElectricPavlov  Oct 26, 2020
 #3
avatar+461 
+1

That makes sense. Thanks a lot!

 Oct 26, 2020
 #4
avatar+461 
+2

Wait actually I obtain the answer of 4. 

HelpBot  Oct 26, 2020
 #5
avatar+111546 
+1

\(ax^2+bx+c=2(x-4)^2+8\\~\\ 3(ax^2+bx+c)=3[2(x-4)^2+8]\\ 3(ax^2+bx+c)=6(x- {\color{red} 4})^2+24\\ \)

SO

h=4

 

 

 

LaTex:

ax^2+bx+c=2(x-4)^2+8\\~\\
3(ax^2+bx+c)=3[2(x-4)^2+8]\\
3(ax^2+bx+c)=6(x -{\color{red} 4})^2+24\\

 Oct 26, 2020
edited by Melody  Oct 26, 2020
 #7
avatar+461 
+1

Yes! That is what I put at first, too, before I made a calculation error and fixed it to 3 sad

 Oct 26, 2020

22 Online Users