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Describe the sequence of transformations that would take the graph of f(x)=x² to each parabola described below. Focus: (1/4,0),directrix: x=-1/4

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Answer:

The parabola obtained is x=y²

Explanation:

Given the focus (1/4,0) and directrix: x=-1/4

A point on the parabola is determined by the distance of that point with the focus and with the directrix:

Distance between the parabola and the directrix:
\sqrt{(x+(1)/(4))^(2)  }

Distance between the parabola and the focus:
\sqrt{(x-(1)/(4))^(2) +(y-0)^(2) }


\sqrt{(x+(1)/(4))^(2)}=
\sqrt{(x-(1)/(4))^(2) +(y-0)^(2) }


(x+(1)/(4))^(2)=
(x-(1)/(4))^(2) +(y-0)^(2)


x^2+(1)/(2) x+(1)/(16) = x^2-(1)/(2) x+(1)/(16) + y^(2)


(1)/(2) x=-(1)/(2) x + y^(2)


(1)/(2) x+(1)/(2) x = y^(2)

x=y²

Describe the sequence of transformations that would take the graph of f(x)=x² to each-example-1
User Honza Hejzl
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