thatis the vertex form of y=2x^2-4+4 with solution po.
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thatis the vertex form of y=2x^2-4+4 with solution po.
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Answer and explation
Question
y=2x
2
−4+4
Rewrite the equation
Rewrite in polar form
r=0
r=
1+cos(2θ)
sin(θ)
Evaluate
y=2x
2
−4+4
Add the terms
y=2x
2
Move the expression to the left side
y−2x
2
=0
To convert the equation to polar coordinates,substitute x for r×cos(θ) and y for r×sin(θ)
sin(θ)r−2(cos(θ)r)
2
=0
Factor the expression
−2cos
2
(θ)r
2
+sin(θ)r=0
Simplify the expression
(−1−cos(2θ))r
2
+sin(θ)r=0
Factor the expression
r((−1−cos(2θ))r+sin(θ))=0
When the product of factors equals 0,at least one factor is 0
r=0
(−1−cos(2θ))r+sin(θ)=0
Solution
r=0
r=
1+cos(2θ)
sin(θ)
Function
Find the vertex
(0,0)
Testing for symmetry
Testing for symmetry about the origin
Not symmetry with respect to the origin
Identify the conic
Write the equation in standard form
x
2
=
2
1
y