Question

How do you write 2z^2+12z-4 in vertex form?

Answer 1

Answer:

2(z + 3)² - 22

Step-by-step explanation:

2z² + 12z - 4

Start by making your "a" value equal to 1 by factoring the 2 from the first 2 terms in the standard form equation.

2(z² + 6z) - 4

Complete the square by using the formula (b/2)². Identify your "b" value, which is 6. Now you can complete the square.

((6)/2)² = 9

After completing the square, add 9 inside the parentheses and subtract 9 outside the parentheses. Since the 9 inside the parentheses is also being multiplied by 2, multiply the subtracted 9 by 2 as well.

2(z² + 6z + 9) - 4 - 9(2)

Factor the terms inside the parentheses by using the product/sum factoring method. This is where you find two of the same terms that multiply to "c" (9) and add to "b" (6).

In this case, positive 3 multiplies to 9 and adds to 6, so we will use the factors (z + 3)(z + 3), which is the same as (z + 3)².

2(z + 3)² - 4 - 9(2)

To finish off the problem, combine the like terms outside of the parentheses by multiplying 9 times 2 first and then subtracting -4 by 9(2).

$\boxed{2(\text{z}+3)^2-22}$