The vertices of cubic function c(x) are located at (−3,1)
 and (2,−9)
 Where are the vertices of 13c(x)
 located? Enter your responses as ordered pairs. Do not include any spaces.

To find the vertices of the function \( 13c(x) \), we need to understand how multiplying a function by a constant affects its vertices. When a function is multiplied by a constant, the vertical positions (y-values) of all points, including the vertices, are multiplied by that constant. The x-values remain unchanged.

The given vertices of \( c(x) \) are:
- \( (-3, 1) \)
- \( (2, -9) \)

For \( 13c(x) \), we multiply the y-coordinates of the vertices by 13:
- For the vertex \( (-3, 1) \), the new y-coordinate is \( 13 \times 1 = 13 \), so the vertex becomes \( (-3, 13) \).
- For the vertex \( (2, -9) \), the new y-coordinate is \( 13 \times (-9) = -117 \), so the vertex becomes \( (2, -117) \).

Thus, the vertices of \( 13c(x) \) are:
- \( (-3, 13) \)
- \( (2, -117) \)

### Final answer: 
\[
(-3,13) \quad \text{and} \quad (2,-117)
\]

Would you like more details or have any questions? 

Here are 5 related questions to expand on this topic:
1. What happens to the x-coordinates of a function when it is multiplied by a constant?
2. How does stretching or compressing a function vertically affect its graph?
3. Can you derive the equation of a cubic function given two of its vertices?
4. What is the difference between vertical scaling and horizontal scaling of a function?
5. How do transformations affect the symmetry of a cubic function?

**Tip:** When a function is scaled by a constant vertically, the y-values change, but the x-values remain the same unless horizontal transformations are involved.

The vertices of cubic function c(x) are located at (−3,1) and (2,−9). Where are the vertices of 13c(x) located? Enter your responses as ordered pairs. Do not include any spaces.

Learn how to find the new vertices of a cubic function after vertical scaling by a constant factor. This problem involves the transformation of the cubic function c(x) where the original vertices are located at (−3,1) and (2,−9). The new vertices of 13c(x) are calculated through vertical stretching. Suitable for students in Grades 10-12.

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