vertices y^2+2y+x+5=0

{ "query": { "display": "vertices $$y^{2}+2y+x+5=0$$", "symbolab_question": "CONIC#vertices y^{2}+2y+x+5=0" }, "solution": { "level": "PERFORMED", "subject": "Geometry", "topic": "Parabola", "subTopic": "vertices", "default": "(h,k)=(-4,-1),p=-\\frac{1}{4}", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "Parabola vertex given $$y^{2}+2y+x+5=0:{\\quad}\\left(h,\\:k\\right)=\\left(-4,\\:-1\\right),\\:p=-\\frac{1}{4}$$", "input": "y^{2}+2y+x+5=0", "steps": [ { "type": "definition", "title": "Parabola standard equation", "text": "$$4p\\left(x-h\\right)=\\left(y-k\\right)^{2}\\:$$ is the standard equation for a right-left facing parabola with vertex at $$\\left(h,\\:k\\right),\\:$$<br/>and a focal length $$|p|$$" }, { "type": "interim", "title": "Rewrite $$y^{2}+2y+x+5=0\\:$$in the standard form", "input": "y^{2}+2y+x+5=0", "steps": [ { "type": "step", "primary": "Rewrite as", "result": "x=-y^{2}-2y-5" }, { "type": "interim", "title": "Complete the square $$-y^{2}-2y-5:{\\quad}-\\left(y+1\\right)^{2}-4$$", "input": "-y^{2}-2y-5", "steps": [ { "type": "step", "primary": "Write $$-y^{2}-2y-5\\:$$in the form: $$x^2+2ax+a^2$$", "secondary": [ "Factor out $$-1$$" ], "result": "-\\left(y^{2}+2y+5\\right)" }, { "type": "interim", "title": "$$2a=2{\\quad:\\quad}a=1$$", "input": "2a=2", "steps": [ { "type": "interim", "title": "Divide both sides by $$2$$", "input": "2a=2", "result": "a=1", "steps": [ { "type": "step", "primary": "Divide both sides by $$2$$", "result": "\\frac{2a}{2}=\\frac{2}{2}" }, { "type": "step", "primary": "Simplify", "result": "a=1" } ], "meta": { "interimType": "Divide Both Sides Specific 1Eq", "gptData": "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" } } ], "meta": { "solvingClass": "Equations", "interimType": "Equations" } }, { "type": "step", "primary": "Add and subtract $$1^{2}\\:$$", "result": "-\\left(y^{2}+2y+5+1^{2}-1^{2}\\right)" }, { "type": "step", "primary": "$$x^2+2ax+a^2=\\left(x+a\\right)^2$$", "secondary": [ "$$y^{2}+2y+1^{2}=\\left(y+1\\right)^{2}$$", "Complete the square" ], "result": "-\\left(\\left(y+1\\right)^{2}+5-1^{2}\\right)" }, { "type": "step", "primary": "Simplify", "result": "-\\left(y+1\\right)^{2}-4" } ], "meta": { "solvingClass": "Equations", "interimType": "Complete Square 1Eq" } }, { "type": "step", "result": "x=-\\left(y+1\\right)^{2}-4" }, { "type": "step", "primary": "Add $$4$$ to both sides", "result": "x+4=-\\left(y+1\\right)^{2}" }, { "type": "step", "primary": "Divide by coefficient of square terms: $$-1$$", "result": "-\\left(x+4\\right)=\\left(y+1\\right)^{2}" }, { "type": "step", "primary": "Rewrite in standard form", "result": "4\\left(-\\frac{1}{4}\\right)\\left(x-\\left(-4\\right)\\right)=\\left(y-\\left(-1\\right)\\right)^{2}" } ], "meta": { "interimType": "Parabola Canonical Format 1Eq" } }, { "type": "step", "result": "4\\left(-\\frac{1}{4}\\right)\\left(x-\\left(-4\\right)\\right)=\\left(y-\\left(-1\\right)\\right)^{2}" }, { "type": "step", "primary": "Therefore parabola properties are:", "result": "\\left(h,\\:k\\right)=\\left(-4,\\:-1\\right),\\:p=-\\frac{1}{4}" } ], "meta": { "solvingClass": "Parabola" } }, "plot_output": { "meta": 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