Cook

Shepherd Center Atlanta, GA
Shepherd Center seeks candidates who are compassionate and caring for patients and their families/caregivers.

Other characteristics we look for include: flexibility, self-motivation and direction, ability to learn and apply knowledge quickly; proficiency in organizing and completing multiple tasks and projects in a timely manner; effective and appropriate communication skills in all interactions; and consistent demonstration of proficiency in area of expertise.

This role is primarily responsible for preparation of quality food in accordance with approved recipes for patient, staff and guests in a sanitary and safe manner.

Responsibilities include: Following standardized recipes and other preparation guidelines in food production, performing simple calculations and adjusting recipes as needed with approval; ensuring items are prepared in a timely manner to meet serving schedule; performing batch and to-order food preparation ensuring good/appealing food quality; practicing sanitary and other guidelines to ensure cleanliness of work area, proper food storage and rotation and minimum waste/overproduction; and working cohesively with managers and team members to deliver overall quality product to patients, visitors and staff.

High school diploma or equivalent required.

Culinary program studies/graduate preferred.

Three years quantity food preparation in healthcare or restaurant and ServSafe certification required.

Full-time position with every other weekend and holiday. Shift schedule will vary.

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