how to obain the formula and the domain of a parabola

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August 3rd 2012, 06:38 PM
wayneB

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how to obain the formula and the domain of a parabola

Finding the formula associated with each parabola?
the question is:
part of a roller coaster is modeled by three parabola - some parts removed to give the diagram below.
The first parabola from the left meets the second at co-ordinates (7, 14). using this information from the diagram how can i find the formula associated with each parabola and also workout their relevant domains? hint: think of the 3rd parabola is a translation of the first.

Here's the diagram:Attachment 24416
after looking at the diagram i plugged the (x, y) into the equation y = ax^2 + bx + c. For each point (x, y).
max(4.5, 20.25)-->y=ax^2+bx+c-->a(4.5)^2+b(4.5)+c
-->20.25a+4.5b+c=14
min(12, 1.5)-->y=ax^2+bx+c-->a(12)^2+b(12)+c
-->144a+12b+c=1.5
max(19.5, 20.25)-->=ax^2+bx+c-->a(19.5)^2+b(19.5)+…
-->380.25a+19.5b+c=20.25
is this correct? what am i meant to do with the co-ordinates (7, 14)? and how do i proceed from this step to obtain not just the formula but ALSO the relevant domains?
August 3rd 2012, 10:36 PM
earboth

Re: how to obain the formula and the domain of a parabola

Quote:

Originally Posted by wayneB

Finding the formula associated with each parabola?
the question is:
part of a roller coaster is modeled by three parabola - some parts removed to give the diagram below.
The first parabola from the left meets the second at co-ordinates (7, 14). using this information from the diagram how can i find the formula associated with each parabola and also workout their relevant domains? hint: think of the 3rd parabola is a translation of the first.

Here's the diagram:Attachment 24416
after looking at the diagram i plugged the (x, y) into the equation y = ax^2 + bx + c. For each point (x, y).
max(4.5, 20.25)-->y=ax^2+bx+c-->a(4.5)^2+b(4.5)+c
-->20.25a+4.5b+c=14
min(12, 1.5)-->y=ax^2+bx+c-->a(12)^2+b(12)+c
-->144a+12b+c=1.5
max(19.5, 20.25)-->=ax^2+bx+c-->a(19.5)^2+b(19.5)+…
-->380.25a+19.5b+c=20.25
is this correct? what am i meant to do with the co-ordinates (7, 14)? and how do i proceed from this step to obtain not just the formula but ALSO the relevant domains?

1. In general your approach to solve this question is OK:

If you want to use the equation of a parabola: $y = ax^2+bx+c$ then you need the coordinates of three points of the parabola. You'll get a system of 3 simultaneous equations with 3 variables ((a,b,c)). Since you know 3 points of each parabola this way will lead to a valid solution. (You certainly noticed that (17, 14) is a point on the 2nd or 3rd parabola)

2. BUT: Since you know the vertex and at least one additional point it would be easier to use the vertex form of the equation of a parabola: If $V(x_V, y_V)$ is the vertex of a parabola it has the equation: $y = a(x-x_V)^2+y_V$

3. The equation of the 1st parabola is then: $y = a \left(x-\frac92\right)^2+\frac{81}4$ . Plug in the coordinates (7, 14):

$14 = a \left(7-\frac92\right)^2+\frac{81}4~\implies~a = -1$

Therefore: $y = (-1) \left(x-\frac92\right)^2+\frac{81}4$

Expand the bracket and collect like terms: $y = -x^2+9x$

4. The two other equations can be obtained in a similar way.
August 4th 2012, 03:20 AM
wayneB

Re: how to obain the formula and the domain of a parabola

Quote:

Originally Posted by earboth

1. In general your approach to solve this question is OK:

If you want to use the equation of a parabola: $y = ax^2+bx+c$ then you need the coordinates of three points of the parabola. You'll get a system of 3 simultaneous equations with 3 variables ((a,b,c)). Since you know 3 points of each parabola this way will lead to a valid solution. (You certainly noticed that (17, 14) is a point on the 2nd or 3rd parabola)

2. BUT: Since you know the vertex and at least one additional point it would be easier to use the vertex form of the equation of a parabola: If $V(x_V, y_V)$ is the vertex of a parabola it has the equation: $y = a(x-x_V)^2+y_V$

3. The equation of the 1st parabola is then: $y = a \left(x-\frac92\right)^2+\frac{81}4$ . Plug in the coordinates (7, 14):

$14 = a \left(7-\frac92\right)^2+\frac{81}4~\implies~a = -1$

Therefore: $y = (-1) \left(x-\frac92\right)^2+\frac{81}4$

Expand the bracket and collect like terms: $y = -x^2+9x$

4. The two other equations can be obtained in a similar way.

could you just explain how you got the -1 in this: [ 14 = a(7-4.5)^2+20.25 ==> x =-1 ]
and how 9x was obtained in this:
y = -(x-4.5)^2+20.25
y = -x^2+9x

is there a step into obtain a or just pluggin values to obtain it? with the 9x was it just 4.5 x 2 = 9 ???
August 4th 2012, 03:52 AM
earboth

Re: how to obain the formula and the domain of a parabola

Quote:

Originally Posted by wayneB

could you just explain how you got the -1 in this: [ 14 = a(7-4.5)^2+20.25 ==> x =-1 ]
and how 9x was obtained in this:
y = -(x-4.5)^2+20.25
y = -x^2+9x

is there a step into obtain a or just pluggin values to obtain it? with the 9x was it just 4.5 x 2 = 9 ???

1.

$14 = a(7-4.5)^2+20.25~\implies~14=6.25a+20.25~\implies~ \\ -6.25=6.25a~\implies~a=-1$

2.

$y = -(x-4.5)^2+20.25~\implies~y=-(x^2-9x+20.25)+20.25~\implies~ \\ y=-x^2+9x$
August 4th 2012, 04:04 PM
wayneB

Re: how to obain the formula and the domain of a parabola

Quote:

Originally Posted by earboth

1.

$14 = a(7-4.5)^2+20.25~\implies~14=6.25a+20.25~\implies~ \\ -6.25=6.25a~\implies~a=-1$

2.

$y = -(x-4.5)^2+20.25~\implies~y=-(x^2-9x+20.25)+20.25~\implies~ \\ y=-x^2+9x$

Thank You so much for your help! :)
August 4th 2012, 04:50 PM
skeeter

Re: how to obain the formula and the domain of a parabola

you've posted this twice ... once in the algebra forum (where it belongs) and now in the trig forum.

http://mathhelpforum.com/algebra/201...-parabola.html

DO NOT DOUBLE POST.