shift f(x)= 1/2 sin(2(x+pi/6))-1

{ "query": { "display": "shift $$f\\left(x\\right)=\\frac{1}{2}\\sin\\left(2\\left(x+\\frac{π}{6}\\right)\\right)-1$$", "symbolab_question": "FUNCTION#shift f(x)=\\frac{1}{2}\\sin(2(x+\\frac{π}{6}))-1" }, "solution": { "level": "PERFORMED", "subject": "Functions & Graphing", "topic": "Functions", "subTopic": "shift", "default": "\\mathrm{Phase}:-\\frac{π}{6},\\:\\mathrm{Vertical}:-1", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "Shift of $$\\frac{1}{2}\\sin\\left(2\\left(x+\\frac{π}{6}\\right)\\right)-1:{\\quad}$$Phase$$:-\\frac{π}{6},\\:\\:$$Vertical$$:-1$$", "steps": [ { "type": "definition", "title": "Shift definition", "text": "For $$f\\left(x\\right)=A{\\cdot}g\\left(Bx-C\\right)+D$$, where $$g\\left(x\\right)$$ is one of the basic trig functions, <br/>$$\\frac{C}{B}$$ is phase shift<br/>$$D$$ is vertical shift" }, { "type": "step", "primary": "$$f\\left(x\\right)=\\frac{1}{2}\\sin\\left(2\\left(x+\\frac{π}{6}\\right)\\right)-1$$<br/>$$g\\left(Bx-C\\right)=\\sin\\left(2\\left(x+\\frac{π}{6}\\right)\\right),\\:B=2,\\:C=-\\frac{π}{3},\\:D=-1$$<br/>Therefore, the phase shift $$\\frac{C}{B}$$ is $$\\frac{-\\frac{π}{3}}{2}$$", "result": "-\\frac{π}{6}" }, { "type": "step", "primary": "The vertical shift $$D$$ is $$-1$$", "result": "-1" } ], "meta": { "solvingClass": "Function Shift" } }, "plot_output": { "meta": { "plotInfo": { "variable": "x", "plotRequest": "\\frac{1}{2}\\sin(2(x+\\frac{π}{6}))-1" }, "showViewLarger": true } }, "meta": { "showVerify": true } }