+0

# Show that when k =8, f(x)>0 for all values of x

0
1012
3
+844

$$f(x)= x^2 - (k+8)x + (8k+1)$$

Show that when k =8, f(x)>0 for all values of x

Oct 20, 2018

#1
+28176
+3

When k = 8 we have $$f(x) = x^2-16x+64+1 \text{ or } f(x) = (x - 8)^2 + 1$$

No matter what the value of x is, (x-8)2 is always zero or positive, Hence f(x) is always greater than zero.

Oct 20, 2018
edited by Alan  Oct 20, 2018
#2
+844
+1

but how do we prove that it is always 0 or positive?

i feel like an explanation is needed for the mark

YEEEEEET  Oct 20, 2018
#3
+28176
+1

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.

Alan  Oct 20, 2018
edited by Alan  Oct 20, 2018