f(x)=x2−(k+8)x+(8k+1)
Show that when k =8, f(x)>0 for all values of x
When k = 8 we have f(x)=x2−16x+64+1 or f(x)=(x−8)2+1
No matter what the value of x is, (x-8)2 is always zero or positive, Hence f(x) is always greater than zero.
but how do we prove that it is always 0 or positive?
i feel like an explanation is needed for the mark
When x = 8 we have (x-8) = 0
For any other value of x we have (x-8) is either positive or negative. If you square a positive number you get a positive number; if you square a negative number you get a positive number.
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