SOLUTION: Find the vertex of each parabola and graph it. 1. y^2+4x-6y= -1 2. y=x^2+4x+5 3. y=-x^2-x+1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex of each parabola and graph it. 1. y^2+4x-6y= -1 2. y=x^2+4x+5 3. y=-x^2-x+1      Log On


   

Question 913580: Find the vertex of each parabola and graph it.
1. y^2+4x-6y= -1
2. y=x^2+4x+5
3. y=-x^2-x+1
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Number 1 has a horizontal axis of symmetry; and the other parabolas are routine. This help will be for number 1.

4x=-y%5E2%2B6y-1
Complete The Square for y.
4x=-1%28y%5E2-6y%2B1%29, use %28-6%2F2%29%5E2=9.
4x=-1%28y%5E2-6y%2B9%2B1-9%29
4x=-1%28%28y-3%29%5E2-8%29
x=-%281%2F4%29%28%28y-3%29%5E2-8%29
highlight%28x=-%281%2F4%29%28y-3%29%5E2%2B2%29, now in standard form for parabola with horizontal axis of symmetry.

This parabola is squashed horizontally, and is concave to the left. Vertex is (2, 3).

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex of each parabola and graph it.
Basic equation of parabola:
1. y^2+4x-6y= -1
complete the square:
(y^2-6y+9)+4x=-1+9
(y-3)^2=-4x+8
(y-3)^2=-4(x-2)
Parabola opens left with vertex at(2,3)
..
2. y=x^2+4x+5
complete the square:
y=(x^2+4x+4)-4+5
(x+2)^2=(y-1)
Parabola opens up with vertex at(-2,1)
..
3. y=-x^2-x+1
complete the square:
y=-(x^2+x+1/4)+1/4+1
(x+1/2)^2=-(y-5/4)
Parabola opens down with vertex at(-1/2,5/4)
Sorry, I don't have the means to graph these parabolas