Question

h) use the equation to complete the table of values
$y = 2(x - 1)^2$

x	y
-1	-6
0
1

Explanation:

Step1: For \( x = 0 \)

Substitute \( x = 0 \) into the equation \( y = 2(x - 1)^2 \).
First, calculate \( x - 1 = 0 - 1=-1 \).
Then, square the result: \( (-1)^2 = 1 \).
Multiply by 2: \( y = 2\times1 = 2 \).

Step2: For \( x = 1 \)

Substitute \( x = 1 \) into the equation \( y = 2(x - 1)^2 \).
Calculate \( x - 1 = 1 - 1 = 0 \).
Square the result: \( 0^2 = 0 \).
Multiply by 2: \( y = 2\times0 = 0 \).
Also, let's re - check \( x=-1 \):
Substitute \( x = - 1 \) into \( y = 2(x - 1)^2 \).
\( x-1=-1 - 1=-2 \), \( (-2)^2 = 4 \), \( y = 2\times4 = 8 \) (the original - 6 is incorrect).

Answer:

For \( x=-1 \), \( y = 8 \); for \( x = 0 \), \( y = 2 \); for \( x = 1 \), \( y = 0 \)

The completed table is:

\( x \)	\( y \)
0	2
1	0