To find the image distance \( v \), we use the mirror equation:\n\[\n\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\n\]\nWhere:\n- \( f = 15 \, \text{cm} \)\n- \( u = -25 \, \text{cm} \)\n\nSubstituting:\n\[\n\frac{1}{15} = \frac{1}{-25} + \frac{1}{v}\n\]\nSolving for \( v \):\n\[\n\frac{1}{v} = \frac{1}{15} + \frac{1}{25} = \frac{5 + 3}{75} = \frac{8}{75}\n\]\n\[\nv = \frac{75}{8} = 9.375 \, \text{cm}\n\]\n\nTherefore, the image forms at \( 9.375 \, \text{cm} \).