Question

completing the square part 1
write the following functions in vertex form by completing the square.

1. $y = x^2 - 6x - 2$

1. $y = x^2 - 4x - 1$

1. $y = x^2 + 2x + 8$

1. $y = x^2 + 10x + 47$

Explanation:

Problem 1: \( y = x^2 - 6x - 2 \)

Step 1: Group \( x \)-terms

\( y=(x^2 - 6x)-2 \)

Step 2: Complete the square for \( x^2 - 6x \)

Take half of \(-6\) (which is \(-3\)), square it (\( (-3)^2 = 9 \)), add and subtract it:
\( y=(x^2 - 6x + 9 - 9)-2 \)

Step 3: Factor the perfect square

\( y=((x - 3)^2 - 9)-2 \)

Step 4: Simplify

\( y=(x - 3)^2 - 9 - 2=(x - 3)^2 - 11 \)

Step 1: Group \( x \)-terms

\( y=(x^2 - 4x)-1 \)

Step 2: Complete the square for \( x^2 - 4x \)

Half of \(-4\) is \(-2\), square is \( (-2)^2 = 4 \), add and subtract:
\( y=(x^2 - 4x + 4 - 4)-1 \)

Step 3: Factor the perfect square

\( y=((x - 2)^2 - 4)-1 \)

Step 4: Simplify

\( y=(x - 2)^2 - 4 - 1=(x - 2)^2 - 5 \)

Step 1: Group \( x \)-terms

\( y=(x^2 + 2x)+8 \)

Step 2: Complete the square for \( x^2 + 2x \)

Half of \( 2 \) is \( 1 \), square is \( 1^2 = 1 \), add and subtract:
\( y=(x^2 + 2x + 1 - 1)+8 \)

Step 3: Factor the perfect square

\( y=((x + 1)^2 - 1)+8 \)

Step 4: Simplify

\( y=(x + 1)^2 - 1 + 8=(x + 1)^2 + 7 \)

Answer:

\( y=(x - 3)^2 - 11 \)

Problem 2: \( y = x^2 - 4x - 1 \)