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August 11th, 2018, 05:42 AM  #1 
Newbie Joined: Aug 2018 From: HK Posts: 17 Thanks: 0  Function and graph
Q: Given that the minimum value of the function y=(x+1)(x3)+k is 8, find the value of k. My answer is 12, however, the model answer is 16. Welcome any idea on this with details calculation, thank you!! 
August 11th, 2018, 06:15 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 
$y = x^22x+(k3)$ $a=1$, $b=2$, $c=k3$ min occurs at $x = \dfrac{b}{2a} = 1 \implies y = k4 = 8 \implies k =12$ 
August 11th, 2018, 05:30 PM  #3 
Newbie Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 
It is a new approach on this kind of questions, thanks a lot

August 11th, 2018, 06:08 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
The formula skeeter used may be derived by completing the square: $$ax^2+bx+c=0$$ $$a\left(x^2+\frac bax\right)+c=0$$ $$a\left(x+\frac{b}{2a}\right)^2+c\frac{b^2}{4a}=0$$ Can you see why the function is at a minimum when $x=\frac{b}{2a}$? Alternatively, to find $k$ solve $c\frac{b^2}{4a}=8$ for $c$, then $k3=c$. 
August 11th, 2018, 07:36 PM  #5 
Newbie Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 
Is this the general form of completing square? Why you need to subtract b^2/4a after c? 
August 12th, 2018, 03:03 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I'm not sure what you mean by "the general form". However, the point of "completing the square" is that a "perfect square" is of the form $\displaystyle (x+ y)^2= x^2+ 2xy+ y^2$. Notice the "2" in the "2xy" term! So if we have something of the form "$\displaystyle ax^2+ bx$" the first thing we can do is factor out that coefficient "a" to get $\displaystyle a(x^2+ (b/a)x)$. Now compare that "$\displaystyle x^2+ (b/a)x$" with the above "$\displaystyle x^2+ 2xy+ y^2$". In order that "(b/a)x" be the same as "2xy" we must have "2y= (b/a)" or "y= b/(2a)". And then $\displaystyle y^2= b^2/(4a^2)$. That is, in order to make "$\displaystyle x^2+ (b/a)x$" a "perfect square" we must add $\displaystyle b^2/(4a)$: $\displaystyle x^2+ (b/a)x+ b^2/4a=(x+ b/(2a))^2$. Of course, if we just add something, we would be changing the value. In order to keep the same value, we must also subtract the same amount $\displaystyle x^2+ (b/a)x= x^2+ (b/a)x+ b^2/(4a^2) b^2/(4a^2)= (x+ b/(2a))^2 b^2/(4a^2)$. Putting that back into the original: $\displaystyle ax^2+ bx= a(x^2+ (b/a)x)= a(x^2+ (b/a)x+ b^2/(4a^2) b^2/(4a^2))= a((x+ b/(2a))^2 b^2/(4a^2))= a(x+ b/(2a))^2 b^2/(4a)$. Last edited by skipjack; August 12th, 2018 at 07:45 AM. 
August 13th, 2018, 10:57 PM  #7 
Newbie Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 
The example of general form (quadratic equation): ax^2+bx+c=0 Thank you for your detail explanation!! 
August 13th, 2018, 11:45 PM  #8 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  

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