Which equation is y = 6x2 + 12x – 10 rewritten in vertex form?
y = 6(x + 1)2 – 11
y = 6(x + 1)2 – 10
y = 6(x + 1)2 – 4
y = 6(x + 1)2 – 16

To rewrite the equation \( y = 6x^2 + 12x - 10 \) in vertex form, we need to complete the square. Let's break this down step-by-step:

### Step 1: Factor out the coefficient of \( x^2 \) (which is 6) from the first two terms.
\[
y = 6(x^2 + 2x) - 10
\]

### Step 2: Complete the square inside the parentheses.
Take half of the coefficient of \( x \) (which is 2), square it, and add it inside the parentheses. So, half of 2 is 1, and \( 1^2 = 1 \). To maintain balance, subtract the same value outside.

\[
y = 6(x^2 + 2x + 1 - 1) - 10
\]
\[
y = 6((x + 1)^2 - 1) - 10
\]

### Step 3: Simplify the equation.
Distribute the 6 and simplify:
\[
y = 6(x + 1)^2 - 6 - 10
\]
\[
y = 6(x + 1)^2 - 16
\]

Thus, the equation in vertex form is:
\[
y = 6(x + 1)^2 - 16
\]

The correct answer is:
\[
\boxed{y = 6(x + 1)^2 - 16}
\]

Would you like more details on completing the square, or any other clarification?

Here are 5 related questions:
1. How do you find the vertex of a quadratic function from its standard form?
2. What are the steps to complete the square in a quadratic equation?
3. How does the vertex form of a quadratic function help in graphing?
4. Can you explain how to factor a quadratic equation in general?
5. What is the significance of the vertex in a parabola?

**Tip:** In vertex form \( y = a(x - h)^2 + k \), the vertex is at \( (h, k) \).

Which equation is y = 6x^2 + 12x – 10 rewritten in vertex form?

Learn how to rewrite the quadratic equation y = 6x^2 + 12x - 10 in vertex form by completing the square. This step-by-step guide explains key algebra concepts like completing the square and finding the vertex of a parabola. Suitable for Grades 9-11.

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