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August 11th, 2018, 06:42 AM   #1
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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!!
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August 11th, 2018, 07:15 AM   #2
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$y = x^2-2x+(k-3)$

$a=1$, $b=-2$, $c=k-3$

min occurs at $x = -\dfrac{b}{2a} = 1 \implies y = k-4 = 8 \implies k =12$
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August 11th, 2018, 06:30 PM   #3
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It is a new approach on this kind of questions, thanks a lot
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August 11th, 2018, 07:08 PM   #4
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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 $k-3=c$.
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August 11th, 2018, 08:36 PM   #5
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Is this the general form of completing square?

Why you need to subtract b^2/4a after c?
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August 12th, 2018, 04:03 AM   #6
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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: $\displaystyle ax^2+ bx= a(x^2+ (b/a)x)= a(x^2+ (b/a)x+ b^2/(4a^2)- b^2/(4a^2))= a((x+ b/(2a))^2- b^2/(4a^2))= a(x+ b/(2a))^2- b^2/(4a)$.
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Last edited by skipjack; August 12th, 2018 at 08:45 AM.
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August 13th, 2018, 11:57 PM   #7
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The example of general form (quadratic equation):

ax^2+bx+c=0


Thank you for your detail explanation!!
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August 14th, 2018, 12:45 AM   #8
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Quote:
Originally Posted by hy2000 View Post
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
$\displaystyle y = a(x - h)^2 + k$

-Dan
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