# Math Help - Maths - Parabola Equations

1. ## Maths - Parabola Equations

Hey guys! I'm new here & browse more thanb I post
I have an emergency, and a lot of guys who are here are pretty knowledgable.
If you could help me out with my question below I would be so happy! Thanks in advance,
Damien.
Question:
I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
I have three points; A (17, 3), B (51,3) and C (34, z)
Z is unkown, however, point C is a local minimum.
So y' = 2ax + b
I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
I have three points; A (17, 3), B (51,3) and C (34, z)
Z is unkown, however, point C is a local minimum.
So y' = 2ax + b. Can anyone help me find the equation of the parabola and the value of z?

2. Hello & welcome

Originally Posted by damien275x
Hey guys! I'm new here & browse more thanb I post
I have an emergency, and a lot of guys who are here are pretty knowledgable.
If you could help me out with my question below I would be so happy! Thanks in advance,
Damien.
Question:
I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
I have three points; A (17, 3), B (51,3) and C (34, z)
Z is unkown, however, point C is a local minimum.
So y' = 2ax + b
I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
I have three points; A (17, 3), B (51,3) and C (34, z)
Z is unkown, however, point C is a local minimum.
So y' = 2ax + b. Can anyone help me find the equation of the parabola and the value of z?
$f(x)=ax^2+bx+c$

The points A, B, C are on the curve.

Therefore, $f(17)=3$, because at the point A, when the absciss is 17, the ordinate is 3.

This goes the same way for B & C : $f(51)=3$ and $f(34)=z$

Substitute in the expression of $f(x)$

These three equations will give you relations between a,b,c, and z.

Now, there is one more information : there is a local minimum at the point C.
This means that at the point C, the derivative is equal to 0.
Therefore $f'(34)=0$.

This will give you a relation between a and b.

So now it's OK, you have 4 equations with 4 unknown variable... Try to get out from this
If you don't understand, just ask...

3. When I did these as simultaneous equations, it kept equating to 0. I was using b = -68a, then trying to find c, with no luck whatsoever. I don't know the relations part.
Could you please go into more detail. I would really appreciate it! Thanks for your help.

4. Originally Posted by damien275x
When I did these as simultaneous equations, it kept equating to 0. I was using b = -68a, then trying to find c, with no luck whatsoever. I don't know the relations part.
Could you please go into more detail. I would really appreciate it! Thanks for your help.
I would prefer you show what you've done ^^
It's easier to explain & point out your mistakes
And I don't see why it would "keep equating to 0"

5. Hi again. I hope you are still there. Not a problem, I will show you what I have come up with so far.
Here it is:

For the form of y = ax^2 + bx +c

For point (17, 3)
3 = a(17^2) + b(17) + c
3 = 289a + 17b + c ... labelling this equation (1)

For point (51, 3)
3 = a(51^2) + b(51) + c
3 = 2601a + 51b + c ... labelling this equation (2)

Stationary points (such as a local minimum) occur when y' = 0
y' = 2ax + b
0 = 2ax + b
0 = 2a(34) + b
0 = 68a + b
b = -68a

Now we have b = -68a
Substitute b = -68a into equation (1)
Therefore, 3 = 289a + 17(-68a) + c
3 = -867a + c
c = 867a + 3

.. and i'm basically stuck after there. Help ! :P

6. you have 4 unknown variables, $a,b,c,z$.. thus you should have 4 equations at first..

use that $f(34) = z$

7. more:

3 = 289a + 17b + c (1)
3 = 2601a + 51b + c (2)

.... (1) x 3 (to eliminate b)
Therefore, 9 = 867a + 51b + 3c (1')

3 = 2601a + 51b + c (2)
9 = 867a + 51b + 3c (1')
(2) - (1')
-6 = 1734a - 2c

f(34) = z
z = a(34^2) + 34b + c
z = 1156a + 34b + c

8. did you substitute that $b = -68a$ to your (1) and (2)?

9. For b = -68a,

For equation (2)
3 = 2601a + 51(-68a) + c
3 = 2601a - 3468a + c
3 = -867a + c

For equation (1)
3 = 289a + 17(-68a) + c
3 = 289a - 1156a + c
3 = -867a + c

Using b = -68a and c = 3 + 867a,
(Substituting these values into equation (1))
3 = 289a + 17(-68a) + 867a + 3
3 = 289a - 1156a + 867a + 3
Which then just gives.. 3 = 0 + 3
3 = 3

10. Okay I have worked everything out:

I have three equations;
289a + 17b + c = 3
2601a + 51b + c = 3
1088a + 33b + c = 0

What do I do now ?

11. Originally Posted by damien275x
Okay I have worked everything out:

I have three equations;
289a + 17b + c = 3
2601a + 51b + c = 3
1088a + 33b + c = 0

What do I do now ?
You're still forgetting this one :

$z=a(34)^2+b(34)+c$

-----------------------------
To sum up :
$f(17)=3 \implies 289a+17b+c=3$

$f(51)=3 \implies 2601a+51b+c=3$

$f(34)=z \implies 1156a+34b+c=z$

$f'(34)=0 \implies 68a+b=0$

From the last one, $b=-68a$.

Substituting in the previous equations, we get :

$\left\{\begin{array}{lll} 289a-1156a+c=3 \\ 2601a-3468a+c=3 \\ 1156a-2312a+c=z \end{array} \right.$

$\left\{\begin{array}{lll} -867a+c=3 \ (1) \\ -867a+c=3 \\ -1156a+c=z \end{array} \right.$

From (1), we get :

$c=3+867a$

Substituting in the last one, we get :

$-1156a+(3+867a)=z$
$\implies z=3-289a$

....

If there is no mistake, there is an infinity of solutions for z !

Why ? because we see that A and B have same ordinate. Plus, a parabola is always symmetric in regard to a line. Here, it would be symmetric in regard to the line x=34, because it's half way from A and B.
If it's symmetric in regard to this line, it means that the minimum is on x=34, which is what we want. But there are many solutions !