Problem Statement

{4,6,8,10,12,14}
6/2
3+1
2*4

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

y = 2x2 - 4x - 6

 
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

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

y = 2x2 - 4x - 6 = 2(x2 - x - )

2
Incorrect
Correct

 
Step 2

Now we can focus on x2 - 2x - 3 and we will remember to multiply it by 2 in Step 4 of this tutorial.

Question Sequence

Question 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 - x +

= (x2 - 2x + ) - 1

= (x - )2 - 1

2
Incorrect
Correct

 
Step 3

Question Sequence

Question 3

Substituting (x − 1)2 − 1 for x2 − 2x in the quadratic x2 − 2x − 3 we have the following.

(x2 - 2x) − 3 = ((x - 1)2 - 1) - 3

= (x - 1)2 -

2
Correct.
Incorrect.

 
Step 4

Question Sequence

Question 4

Now, substituting from Steps 1 to 3 into the original quadratic y = 2x2 − 4x − 6, the completed square form of the quadratic function in the form y = a(x ± h)2 + k is

y = 2x2 − 4x −

= 2(2x2 − 2x − )

= 2((x - 1)2 − )

= 2(x - 1)2 −

2
Correct.
Incorrect.

 
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 5

The term 2(x - 1)2 0 for all x. Thus 2(x - 1)2-8 -8 for all x.

2
Correct.
Incorrect.

 
Step 6

Question Sequence

Question 6

Since 2(x - 1)2-8 ≥ -8, or simply y ≥ -8, the value of y is at x = .

2
Correct.
Incorrect.