Question

completing the square part 2
write the following functions in vertex form by completing the square.

1. $y = -3x^2 + 18x - 30$

1. $y = 3x^2 - 6x - 15$

1. $y = -2x^2 - 16x + 4$

1. $y = 5x^2 + 20x + 7$

Explanation:

Problem 1: \( y = -3x^2 + 18x - 30 \)

Step 1: Factor out the coefficient of \( x^2 \) from the first two terms.

\( y = -3(x^2 - 6x) - 30 \)

Step 2: Complete the square inside the parentheses. The coefficient of \( x \) is -6, half of it is -3, and squaring it gives 9. So we add and subtract 9 inside the parentheses. But since there's a factor of -3 outside, we actually add \( -3 \times 9 \) to the outside.

\( y = -3(x^2 - 6x + 9 - 9) - 30 \)
\( y = -3((x - 3)^2 - 9) - 30 \)

Step 3: Distribute the -3.

\( y = -3(x - 3)^2 + 27 - 30 \)

Step 4: Simplify the constants.

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

Step 1: Factor out the coefficient of \( x^2 \) from the first two terms.

\( y = 3(x^2 - 2x) - 15 \)

Step 2: Complete the square inside the parentheses. The coefficient of \( x \) is -2, half of it is -1, and squaring it gives 1. So we add and subtract 1 inside the parentheses. Then add \( 3 \times 1 \) to the outside (because of the factor 3).

\( y = 3(x^2 - 2x + 1 - 1) - 15 \)
\( y = 3((x - 1)^2 - 1) - 15 \)

Step 3: Distribute the 3.

\( y = 3(x - 1)^2 - 3 - 15 \)

Step 4: Simplify the constants.

\( y = 3(x - 1)^2 - 18 \)

Step 1: Factor out the coefficient of \( x^2 \) from the first two terms.

\( y = -2(x^2 + 8x) + 4 \)

Step 2: Complete the square inside the parentheses. The coefficient of \( x \) is 8, half of it is 4, and squaring it gives 16. So we add and subtract 16 inside the parentheses. Then add \( -2 \times (-16) \) to the outside (because of the factor -2).

\( y = -2(x^2 + 8x + 16 - 16) + 4 \)
\( y = -2((x + 4)^2 - 16) + 4 \)

Step 3: Distribute the -2.

\( y = -2(x + 4)^2 + 32 + 4 \)

Step 4: Simplify the constants.

\( y = -2(x + 4)^2 + 36 \)

Answer:

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

Problem 2: \( y = 3x^2 - 6x - 15 \)