+0  
 
0
79
3
avatar+324 

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

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

YEEEEEET  Oct 20, 2018
 #1
avatar+27128 
+2

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.

Alan  Oct 20, 2018
edited by Alan  Oct 20, 2018
 #2
avatar+324 
+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
avatar+27128 
+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

30 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.