SOLUTION: Please help me solve this equation Find the Parabola properties and graph x^2+4x+2y=0 Vertex: Focus: Directrix: Axis of symmetry: Endpoints of latus retum:

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me solve this equation Find the Parabola properties and graph x^2+4x+2y=0 Vertex: Focus: Directrix: Axis of symmetry: Endpoints of latus retum:       Log On


   

Question 1044783: Please help me solve this equation
Find the Parabola properties and graph
x^2+4x+2y=0
Vertex:
Focus:
Directrix:
Axis of symmetry:
Endpoints of latus retum:

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 +4x +2y = 0
:
y = (-x^2/2) -2x
:
This is a parabola that curves downward
:
+graph%28+300%2C+200%2C+-5%2C+1%2C+-2%2C+3%2C+%28-x%5E2%2F2%29+-2x%29+
:
the x coordinate of the vertex is defined
:
x = -b/2a = -(-2) / (2(-1/2)) = -2
:
y = (-2^2/2) -2(-2) = 2
:
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vertex is (-2,2)
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:
y = (-x^2/2) -2x
:
2y = -x^2 -4x
:
-2y = x^2 +4x
:
Complete the square
:
-2y +4 = (x+2)^2
;
This is the standard form equation for our parabola
:
-2(y-2) = (x+2)^2
:
note that 4p = -2, h = -2, k = 2
:
4p = -2
:
p = -1/2
:
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Focus = (-2, 2-(1/2)) = (-2, 1.5)
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:
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Directrix = 2-(-1/2) = 2.5
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:
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axis of symmetry is x = -2
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: