To convert the given function into standard form and identify its components, let's analyze the function step-by-step.

The function given is: $y = 6 \left( x - \frac{1}{6} \right)^2 + 6$

Step 1: Expand the function

To express this in standard form, $y = ax^2 + bx + c$, we need to expand the expression.

1. Expand $\left( x - \frac{1}{6} \right)^2$: $\left( x - \frac{1}{6} \right)^2 = x^2 - \frac{2}{6}x + \frac{1}{36} = x^2 - \frac{1}{3}x + \frac{1}{36}$

2. Now, multiply each term by 6: $y = 6 \left( x^2 - \frac{1}{3}x + \frac{1}{36} \right) + 6$ $= 6x^2 - 2x + \frac{6}{36} + 6$ $= 6x^2 - 2x + \frac{1}{6} + 6$

3. Combine the constants: $y = 6x^2 - 2x + \frac{37}{6}$

So, the function in standard form is: $y = 6x^2 - 2x + \frac{37}{6}$

Step 2: Identify the Vertex

Since the function was originally in vertex form $y = a(x - h)^2 + k$, we can identify:

• $h = \frac{1}{6}$
• $k = 6$

Thus, the vertex is at: $\left( \frac{1}{6}, 6 \right)$

Step 3: Identify the Stretch/Compression Factor

The coefficient of the $(x - h)^2$ term is 6, indicating a vertical stretch by a factor of 6.

Summary of Answers

• Function in standard form: $y = 6x^2 - 2x + \frac{37}{6}$
• Vertex: $\left( \frac{1}{6}, 6 \right)$
• Stretch/compression factor: 6

Would you like further details on any part of this solution, or do you have other questions?

Here are some related questions:

1. How do we convert from vertex form to standard form for any quadratic function?
2. Why does a coefficient greater than 1 indicate a stretch rather than a compression?
3. How can we find the vertex of a quadratic function given in standard form?
4. What is the effect of negative values for the stretch/compression factor?
5. How does changing the vertex affect the graph of a quadratic function?

Tip: Converting functions between vertex form and standard form is helpful for graphing because it reveals both the vertex and the shape (stretch/compression) of the parabola.