Chapter 1.

1.1 Problem Statement

{4,6,8,10,12,14}
$c/2
$cdiv2+1
2*$cdiv2plus1

Complete the square and find the minimum or maximum value of the quadratic function.

y = 2x2 - 4x - $c

1.2 Step 1

Completing the square of a quadratic function, y = ax2 ± bx + c, means to edit its form by appropriately adding and subtracting a constant so that the function contains a perfect square trinomial and can be rewritten as y = a(x ± h)2 + k, where a, h and k are constants. There are multiple approaches to accomplishing this.

Question Sequence

Question 1.1

The approach of this tutorial will begin by factoring out the leading coefficient, the coefficient of x2, to make it 1.

y = 2x2 - 4x - $c = 2(x2 - XvVM00l89Is=x - 2m8LMWO29ieZby3A)

2
Incorrect
Correct

1.3 Step 2

Now we can focus on x2 - 2x - $cdiv2 and we will remember to multiply it by 2 in Step 4 of this tutorial.

Question Sequence

Question 1.2

We want to complete the square for the terms x2 − 2x by adding and subtracting the square of half the coefficient of x, , and grouping the first three terms which forms a perfect square trinomial.

x2 - 2x = x2 - XvVM00l89Is=x +

= (x2 - 2x + 0VV1JcqyBrI=) - 1

= (x - 0VV1JcqyBrI=)2 - 1

2
Incorrect
Correct

1.4 Step 3

Question Sequence

Question 1.3

Substituting (x − 1)2 − 1 for x2 − 2x in the quadratic x2 − 2x − $cdiv2 we have the following.

(x2 - 2x) − $cdiv2 = ((x - 1)2 - 1) - $cdiv2

= (x - 1)2 - jajrMk3PoV7AciqBQMDtEA==

2
Correct.
Incorrect.

1.5 Step 4

Question Sequence

Question 1.4

Now, substituting from Steps 1 to 3 into the original quadratic y = 2x2 − 4x − $c, the completed square form of the quadratic function in the form y = a(x ± h)2 + k is

y = 2x2 − 4x − SFgqQUkJGdg=

= 2(2x2 − 2x − 2m8LMWO29ieZby3A)

= 2((x - 1)2 − jajrMk3PoV7AciqBQMDtEA==)

= 2(x - 1)2 − pdjJ+iAU87jvS8mqm1FimztGaJPzTZEy

2
Correct.
Incorrect.

1.6 Step 5

The maximum value (or minimum value) of a function is the greatest (or least) value of y = f(x) for any value of x in the domain of the function.

Question Sequence

Question 1.5

The term 2(x - 1)2WLwv7gONmdTuBoqL/S40/M2xUdssSlsqcmmkfIYB8q/ziZof/nqwCawAHmKzYxeH 0 for all x. Thus 2(x - 1)2-$cdiv2plus1times2 WLwv7gONmdTuBoqL/S40/M2xUdssSlsqcmmkfIYB8q/ziZof/nqwCawAHmKzYxeH -$cdiv2plus1times2 for all x.

2
Correct.
Incorrect.

1.7 Step 6

Question Sequence

Question 1.6

Since 2(x - 1)2-$cdiv2plus1times2 ≥ -$cdiv2plus1times2, or simply y ≥ -$cdiv2plus1times2, the iNI7R+IXqcLnJfkJtmHHDPtsRW8= value of y is pdjJ+iAU87jvS8mqm1FimztGaJPzTZEy at x = 0VV1JcqyBrI=.

2
Correct.
Incorrect.