Question

Rewrite each relation in the form y = a(x - h)^2 + k by completing the square

Answer 1

The
`\( y = (x + 3)^2 - 10 \)` is the quadratic equation in the desired form.

The
`\( y = (x - 3)^2 - 13 \)` is the quadratic equation in the desired form.

a)
`\( y = x^2 + 6x - 1 \)`:

To rewrite this quadratic equation in the form
`\(y = a(x - h)^2 + k\)`, we complete the square:

`\[ y = x^2 + 6x - 1 \]`

1. **Group the x-terms:**

`\[ y = (x^2 + 6x) - 1 \]`

2. **Complete the square (add and subtract
`\((6/2)^2 = 9\))`:**

`\[ y = (x^2 + 6x + 9 - 9) - 1 \]`

3. **Factor the perfect square trinomial and simplify:**

`\[ y = (x + 3)^2 - 10 \]`

So,
`\( y = (x + 3)^2 - 10 \)` is the quadratic equation in the desired form.

b.
`\( y = x^2 - 6x - 4 \)`:

To rewrite this quadratic equation in the form
`\(y = a(x - h)^2 + k\)`, we complete the square:

`\[ y = x^2 - 6x - 4 \]`

1. **Group the x-terms:**

`\[ y = (x^2 - 6x) - 4 \]`

2. **Complete the square (add and subtract
`\((-6/2)^2 = 9\))`:**

`\[ y = (x^2 - 6x + 9 - 9) - 4 \]`

3. **Factor the perfect square trinomial and simplify:**

`\[ y = (x - 3)^2 - 13 \]`

So,
`\( y = (x - 3)^2 - 13 \)` is the quadratic equation in the desired form.

Answer 2

Use:
`(*)\ (a\pm b)^2=a^2\pm2ab+b^2`

`a)\\y=x^2+6x-1=\underbrace{x^2+2\cdot x\cdot3+3^2}_((*))-3^2-1=(x+3)^2-9-1=(x+3)^2-10`



`b)\\y=x^2-6x-4=\underbrace{x^2-2\cdot x\cdot3+3^2}_((*))-3^2-4=(x-3)^2-9-4=(x-3)^2-13`