The function y = 4x^2+2x+6 undergoes a horizontal stretch by a factor of 1/2 about the y-axis. The resulting function is 

When a function undergoes a horizontal stretch by a factor of \( \frac{1}{2} \) about the y-axis, it affects the \( x \)-coordinate of the function. This transformation is done by replacing \( x \) with \( \frac{x}{k} \) in the original function, where \( k \) is the stretch factor. In this case, the factor is \( \frac{1}{2} \), so we replace \( x \) with \( \frac{x}{\frac{1}{2}} = 2x \).

Given the original function:

\[
y = 4x^2 + 2x + 6
\]

The horizontal stretch by a factor of \( \frac{1}{2} \) gives:

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

Now, let's simplify this:

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

\[
y = 16x^2 + 4x + 6
\]

So, the resulting function after the horizontal stretch is:

\[
y = 16x^2 + 4x + 6
\]

### Would you like more details or have any questions about this transformation?

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Here are 5 related questions:

1. How does a vertical stretch differ from a horizontal stretch in transformations?
2. What happens if the stretch factor was \(2\) instead of \( \frac{1}{2} \)?
3. How does a reflection about the y-axis affect the function?
4. Can you perform a horizontal translation combined with the stretch? How?
5. What are the effects of stretching the function along both axes simultaneously?

**Tip:** When stretching a function horizontally, the stretch factor affects only the \(x\)-variable, causing the graph to either compress or expand along the x-axis.

The function y = 4x^2+2x+6 undergoes a horizontal stretch by a factor of 1/2 about the y-axis. The resulting function is

Discover how to apply a horizontal stretch by a factor of 1/2 to the quadratic function y = 4x^2 + 2x + 6. This step-by-step solution explores transformations in algebra, focusing on how to manipulate quadratic functions. Ideal for students in Grades 9-12.

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