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Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point Author: QUITTNER [ Next Thread | Previous Thread | Next Message | Previous Message ] Date Posted: 10:17:49 09/09/04 Thu In reply to: masnah binti abd.rahim 's message, "Re: Finding the Equation of a Parabola using the vertex and 1 point" on 02:59:18 09/09/04 Thu masnah binti abd.rahim, WHAT IS YOUR QUESTION? [ Next Thread | Previous Thread | Next Message | Previous Message ]

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[> [> [> Subject: Re: Finding the Equation of a Parabola Author: Justin [Edit] Date Posted: 15:55:44 02/21/05 Mon if the problem says express each equation in the form y=a(x-h)²+k or x=a(y-k)²+h, then how do you do y=2x²-12x+6? [ Post a Reply to This Message ]

[> [> [> [> Subject: Re: Finding the Equation of a Parabola Author: QUITTNER [Edit] Date Posted: 10:00:46 02/22/05 Tue >>> if the problem says express each equation in the form y=a(x-h)²+k or x=a(y-k)²+h, then how do you do y=2x²-12x+6? ..... First you selcet the proper equation. Here the left side is y, therefore use y=a(x-h)^2+k. Expand this to get y=a*(x^2-2hx+h^2) + k or y=a(x^2) - 2ahx +a(h^2) + k (1) We also know y=2x^2) - 12x + 6 (2) Compare the coeficients of equation (1) with those of (2): a=2 4h=12, h=3 18+k=6, k=-12 Always check your answers: y=2*(x-3)^2 - 12 = 2*(x^2-6x+9) - 12 = y=2*(x^2)-12x+18-12 OK [ Post a Reply to This Message ]