# Thread: how to obain the formula and the domain of a parabola

1. ## 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:

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?

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

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:

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: $\displaystyle y = ax^2+bx+c$ 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 $\displaystyle V(x_V, y_V)$ is the vertex of a parabola it has the equation: $\displaystyle y = a(x-x_V)^2+y_V$

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

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

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

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

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

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

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: $\displaystyle y = ax^2+bx+c$ 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 $\displaystyle V(x_V, y_V)$ is the vertex of a parabola it has the equation: $\displaystyle y = a(x-x_V)^2+y_V$

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

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

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

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

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 ???

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

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.

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

2.

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

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

Originally Posted by earboth
1.

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

2.

$\displaystyle y = -(x-4.5)^2+20.25~\implies~y=-(x^2-9x+20.25)+20.25~\implies~ \\ y=-x^2+9x$
Thank You so much for your help!

6. ## 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.

Find the Formula Associated with Each parabola

DO NOT DOUBLE POST.