So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
G(x)=8(x-3)-7
The number in the parentheses has to be changed to positive 3.
You keep the number (-7) the same sign it already is. The vertex is (3,-7)
You transformed the function as follows:
All trasformations of the fashion 
translate the function horizontally, k units right if k is positive, k units left if k is negative.
In your case, k=6, so you translated the original function y=|x| 6 units to the right.
Quadratic Formula is (-b ±√(b²-4ac))/2a
With that being said:
y=2x²-5x-1 is in the correct form of ax²+bx+c
Now just plug in:
5±√(25-4(2 x-1))
5± √25+8
(5±√33)/2(2)
Answer is D
100% Sure
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