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a,b, and c are constants such that the quadratic ax^2 + bx + c can be expressed in the form 2(x - 4)^2 + 8. When the quadratic 3ax^2 + 3bx + 3c (for the same values of a, b, and c) is expressed in the form n(x - h)^2 + k, what is h?

 

thanks you very much!!!!!!!!!!

HELP PLEASE NOW

somebody  Oct 19, 2018
 #1
avatar+2729 
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\(2(x-4)^2 + 8 = 2(x^2 - 8x+16)+8 = 2x^2-16x+32+8 = 2x^2-16x+40 \\ a=2,~b=-16,~c=40\\ 3ax^2+3bx+3c = 6x^2-48x+120\\ 6(x^2-8x+20) = 6((x-4)^2+4) = 6(x-4)^2 + 24 \\ h=4\\ \text{Having gone through all that it's pretty clear to see that this should be so }\\ \text{since multiplying all the constants by 3 just has the effect of scaling the parabola}\\ \text{it's location remains unchanged. The x coordinate of the vertex of the original parabola}\\ \text{was 4, and thus we expect that the scaled parabola's vertex also has an x coordinate of 4}\)

Rom  Oct 19, 2018

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