Use the graphs of  $f$f​  and $h$h​  to describe the transformation from the graph of  $f$f​  to the graph of $h$h​ .

$f\left(x\right)=3x-12$f(x)=3x−12​

$h\left(x\right)=\frac{1}{6}f\left(x\right)$h(x)=16​f(x)​

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The graph of $h$h​  is a Response area of the graph of  $f$f​  by a factor of Response area.

The graph of $ h(x) $ is a **vertical compression** of the graph of $ f(x) $ by a factor of **6**.

This means that each $ y $-value of $ f(x) $ is multiplied by $ \frac{1}{6} $ to produce $ h(x) $, which brings the graph closer to the $ x $-axis without affecting the $ x $-values.

Would you like more details on transformations or have any questions?

1. How do vertical compressions affect the steepness of linear graphs?
2. What is the difference between vertical and horizontal compressions in graph transformations?
3. How can you determine the transformation factor by comparing functions directly?
4. How does changing the coefficient of a function affect its intercepts?
5. How would this transformation look if we had a quadratic or another nonlinear function?

**Tip:** For any transformation of $ y = f(x) $, a vertical compression is achieved by multiplying $ f(x) $ by a factor between 0 and 1.

Use the graphs of f and h to describe the transformation from the graph of f to the graph of h. f(x) = 3x - 12, h(x) = (1/6)f(x)

Explore how to describe the transformation from the graph of f(x) = 3x - 12 to h(x) = (1/6)f(x). This math problem explains vertical compression in linear transformations, with examples suitable for Grades 9-10.

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