Finding the vertex of a parabola from an equation

  • Thread starter DLxX
  • Start date
  • #1
58
0
I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.
 

Answers and Replies

  • #2
319
0
DLxX said:
I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.

by differentiating ? or the line of symmetry between the roots of the equation ?



or completin the square ?
 
Last edited:
  • #3
789
4
Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)
 
  • #4
58
0
Jameson said:
Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)
would be a sideways since the y is squared, but how would i find the h in (h,k)?
 
  • #5
789
4
Show me the equation in the form of x =. Start from there. Did you solve for k?
 
  • #6
58
0
Jameson said:
Show me the equation in the form of x =. Start from there. Did you solve for k?
x = y^2 + 4y + k. So to solve for k do I then plug in 12 and 1 for x and y?
 
  • #7
58
0
DeathKnight said:
Yup thats what you have to do.
Alright so that would give me the k value, but what about the h value?
 
  • #8
319
0
DLxX said:
I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.
plug in values , then just complete the square to put it in the form

f(y) = x = (y+a)^2+b

and b,-a is the vertex
 

Related Threads on Finding the vertex of a parabola from an equation

Replies
3
Views
18K
Replies
3
Views
4K
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
561
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
9K
Top