How to find the equation of a parabola with the following vertex.

  • Thread starter Thread starter Cobalt
  • Start date Start date
  • Tags Tags
    Parabola Vertex
AI Thread Summary
To find the equation of a parabola with a vertex at (-3, -2) that opens downward, the standard form is y = a(x + 3)² - 2, where 'a' is negative. The initial attempt of y = -3x² - 3x - 2 was incorrect. The correct approach involves using the vertex form and ensuring 'a' is a negative value to achieve the downward opening. The discussion emphasizes the importance of correctly identifying the vertex and the sign of 'a' in the equation. Understanding this will help in accurately determining the parabola's equation.
Cobalt
Messages
1
Reaction score
0
1. I need to find the equation of a parabola with a vertex of (-3,-2). It has to open down.



2. I understand of hopefully will understand it.



3. My answer was y=-3x Squared -3x-2 Which I found to be wrong.

 
Physics news on Phys.org
Welcome to PF!

Hi Cobalt! Welcome to PF! :smile:
Cobalt said:
1. I need to find the equation of a parabola with a vertex of (-3,-2). It has to open down.

My answer was y=-3x Squared -3x-2 Which I found to be wrong.

yes, it goes through (-3,-2) …

but it's -3(x + 0.5)2 + something. :wink:
 
A parabola with vertex at (x0, y0) is given by
y= a(x- x0)2+ y0. If opens downward, then a must be negative.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top