1. **State the problem:** We need to find the image of triangle \(\triangle STU\) after a dilation centered at the origin with a scale factor of \(\frac{1}{4}\). 2. **Recall the dilation formula:** For a point \((x,y)\), dilation centered at the origin with scale factor \(k\) maps it to \((kx, ky)\). 3. **Apply the dilation to each vertex:** - \(S(-6, -8) \to S'\left(\frac{1}{4} \times -6, \frac{1}{4} \times -8\right) = \left(-\frac{6}{4}, -\frac{8}{4}\right) = \left(-\frac{3}{2}, -2\right)\) - \(T(8, -8) \to T'\left(\frac{1}{4} \times 8, \frac{1}{4} \times -8\right) = \left(2, -2\right)\) - \(U(-8, 4) \to U'\left(\frac{1}{4} \times -8, \frac{1}{4} \times 4\right) = \left(-2, 1\right)\) 4. **Final answer:** The image \(\triangle S'T'U'\) has vertices at \(S'\left(-\frac{3}{2}, -2\right)\), \(T'(2, -2)\), and \(U'(-2, 1)\).