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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)(x-3)+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: 3,018 Thanks: 1604 $y = x^2-2x+(k-3)$ $a=1$, $b=-2$, $c=k-3$ min occurs at $x = -\dfrac{b}{2a} = 1 \implies y = k-4 = 8 \implies k =12$ Thanks from greg1313 and hy2000 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: 7,964 Thanks: 1148 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 $k-3=c$. Thanks from topsquark and hy2000 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: $\displaystyle ax^2+ bx= a(x^2+ (b/a)x)= a(x^2+ (b/a)x+ b^2/(4a^2)- b^2/(4a^2))= a((x+ b/(2a))^2- b^2/(4a^2))= a(x+ b/(2a))^2- b^2/(4a)$. Thanks from greg1313 and topsquark 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
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 Originally Posted by hy2000 The example of general form (quadratic equation): ax^2+bx+c=0 Thank you for your detail explanation!!
Everyone instructor has their favorite. One of the other typical forms is
$\displaystyle y = a(x - h)^2 + k$

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