Question 160957
I think the procedure is just start at a point get a value for the function, move over get another point, etc. until you see the pattern that the function has either maxed out (reached a maximum) or bottomed out (reached a minimum). Plug and chug we used to call it.
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a)
y=x^2-4x+1
y(0)=0-0+1=1
y(1)=1-4+1=-2
y(2)=4-8+1=-3
y(3)=9-12+1=-2
Looks like (2,-3) is the minimum.
{{{ graph( 300, 300, -5, 5, -5, 5, x^2-4x+1) }}} 
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b)
y=x^2-8x-5
y(2)=4-16-5=-17
y(3)=9-24-5=-20
y(4)=16-32-5=-21
y(5)=25-40-5=-20
Looks like (4,-21) is the minimum.
{{{ graph( 300, 300, -2, 7, -25, 5, x^2-8x-5) }}} 
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c)
y=-2x^2+2x+6
y(-1)=-2-2+6=2
y(0)=0+0+6=6
y(1)=-2+2+6=6
y(2)=-8+4+6=2
The maximum is between 0 and 1, at x=1/2.
y(1/2)=-1/2+1+6=6.5
(0.5,6.5)
{{{ graph( 300, 300, -5, 5, -2, 8, -2x^2+2x+6) }}}
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Just a hint, for quadratic equations in the form {{{ax^2+bx+c}}}, the max or min always occurs at
{{{x=-b/2a}}}
so if we look at your examples again,

y=x^2-4x+1, a=1, b=-4, {{{x=-(-4)/(2(1))=2}}}

y=x^2-8x-5, a=1, b=-8, {{{x=-(-8)/(2(1))=4}}}

y=-2x^2+2x+6, a=-2, b=2, {{{x=-(2)/(2(-2))=1/2}}}
Same answers just much faster than guessing.