another maths question

mattheman

mattheman

mattheman

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13 Oct 2008
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complete the square

y=2x^2-3x-1

y= 2{ x^2-3/2x}-1
x=(3/4) (/16

y=2{x^-3/2 x+ 9/16}-1-9/8
y=2(x-3/4)^2 -17/8

x=3/4 min point y = -17/8


but i did it again and it gave me

2{x^2(3/2)^2 x -(1/2)^2 +(3/2)^2

2(x-3/2)^2 -1/2 -9/4

2(x-3/2)^2 -2/4 -9/4=11/4

x=3/2 y=2x11= 22/4= 11/2
 
It's a real pain in the arse trying to decipher equations written like that so i just did the problem instead and you can see for yourself where you went wrong :)

y= 2x² - 3x - 1
= 2 (x - 3/4)² - 1 - 2(-3/4)²
= 2 (x - 3/4)² - 17/8

So the lowest point is at y = -17/8, x = 3/4. Which i see is what you got first time round, so whatever you did 2nd time is wrong
 
Ugh, completing the square? What a horrible method. Differentiate!

y = 2x^2 - 3x - 1

dy/dx = 4x - 3

dy/dx = 0 implies x = 3/4

Plug into original: y = -17/8
 
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