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Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
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Date Posted: 09:11:49 11/10/03 Mon
In reply to: Ed 's message, "Finding the Equation of a Parabola using the vertex and 1 point" on 22:13:15 11/05/03 Wed
 >vertex (5,12), Point (7,15)
>I know that standard form is f(x)=a(x-h)^2+k
>how do i plug these in to get the proper answer.
>The book gives the answer as -3/4(x-5)^2+12
..... We have two points (x,y), with the vertex being one of them. We also know that
y=a(x-h)^2+k This is (y-k)=a*(x-h)^2
Note that we deal here with the differences in x and in y, distances from the (given) vertex. Therefore k=the y of the vertex, and h is the x of the vertex.
Therefore (15-12)=a*(7-5)^2
3=a*(2)^2 and 3=a*4 giving a=3/4
The parabola's equation is therefore for ANY x and y:
y=(3/4)*(x-5)^2 + 12 COMPARE with the book's answer.
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Replies:
[> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
masnah binti abd.rahim
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Date Posted: 02:59:18 09/09/04 Thu
>>vertex (5,12), Point (7,15)
>>I know that standard form is f(x)=a(x-h)^2+k
>>how do i plug these in to get the proper answer.
>>The book gives the answer as -3/4(x-5)^2+12
>..... We have two points (x,y), with the vertex being
>one of them. We also know that
>y=a(x-h)^2+k This is (y-k)=a*(x-h)^2
>Note that we deal here with the differences in x and
>in y, distances from the (given) vertex. Therefore
>k=the y of the vertex, and h is the x of the vertex.
>Therefore (15-12)=a*(7-5)^2
>3=a*(2)^2 and 3=a*4 giving a=3/4
>The parabola's equation is therefore for ANY x and y:
>y=(3/4)*(x-5)^2 + 12 COMPARE with the book's answer.

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
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Date Posted: 10:17:49 09/09/04 Thu
masnah binti abd.rahim, WHAT IS YOUR QUESTION?

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[> [> [> Subject: Re: Finding the Equation of a Parabola


Author:
Justin
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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?

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[> [> [> [> Subject: Re: Finding the Equation of a Parabola


Author:
QUITTNER
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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

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[> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
Kristen
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Date Posted: 09:33:14 10/18/04 Mon
y=a(h-h)* +k (*=^2)
fill in your points
15=a(7-5)*+12
15-12=a(2)*
3=4a
a=3/4

Standard, fill in a and vertex
a(x-h)*+k
3/4(x-5)*+12

thus that is your equation in standard form

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
Kristen
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Date Posted: 09:34:19 10/18/04 Mon
>y=a(h-h)* +k (*=^2)
>fill in your points
>15=a(7-5)*+12
>15-12=a(2)*
>3=4a
>a=3/4
>
>Standard, fill in a and vertex
>a(x-h)*+k
>3/4(x-5)*+12
>
>thus that is your equation in standard form

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
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Date Posted: 07:49:56 10/19/04 Tue
Given that the vertex has coordinates (a,b), and the one give point has coordinats (x1,y1)
Starting with the equation of a parabola:
(y-a)^2 = k * (x-b)
We know a and b (the vertex), and we can then use
(y1-a)^2 = k * (x1-b) to find k

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[> [> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
QUITTNER
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Date Posted: 08:05:59 10/19/04 Tue
CORRECTION: (Reversing a and b in my previous post):

Given that the vertex has coordinates (a,b), and the
one given point has coordinates (x1,y1)

Starting with the equation of a parabola:
(y-b)^2 = k * (x-a)
We know a and b (the vertex), and we can then use
(y1-b)^2 = k * (x1-a) to find k

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[> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
leslie ann peji
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Date Posted: 03:48:04 08/16/10 Mon
the x and y coordinates of the vertex gives the values of h and k respectively
hence, h=5 and k=12

the equation can be written as y=a(x-5)^2+12 "a" can be found using the fact that the point (7, 15)is on the graph of the line. 15= a(7-5)^2+12
15=4a+12
4a=15-12
a=-3/4
therefore, y=a(x-h)^2+12

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
leslie ann peji
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Date Posted: 03:54:36 08/16/10 Mon
>the x and y coordinates of the vertex gives the values
>of h and k respectively
> hence, h=5 and k=12
>
>the equation can be written as y=a(x-5)^2+12 "a" can
>be found using the fact that the point (7, 15)is on
>the graph of the line. 15= a(7-5)^2+12
> 15=4a+12
> 4a=15-12
> a=-3/4
> therefore, y=a(x-h)^2+12 answer: y=-3/4(x-5)^2+12

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[> [> [> Subject: Re: Finding the Equation of a Parabola using the vertex and 1 point


Author:
simon (sad)
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Date Posted: 05:56:35 03/23/13 Sat
I am finding it hard to find the equation of the parabola given the vertex(7,-2) and a x-intercept(4,0). How will i get or find the equation of the parabola.

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