Solve y=2x^2-16x^33

Function

Evaluate the derivative

Find the domain

Find the x -intercept/zero

y^{'} = 4 x - 144 x^{2}

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Testing for symmetry about the origin

Testing for symmetry about the x-axis

Testing for symmetry about the y-axis

Not symmetry with respect to the origin

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y = 2 x^{2} - 48 x^{3}

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r = 0 r = \frac{cos ^{2} ( θ ) + cos ( θ ) ( cos ( θ ) - 48 sin ( θ ) ) \times ∣ cos ( θ ) ∣}{48 cos ^{3} ( θ )} r = \frac{cos ^{2} ( θ ) - cos ( θ ) ( cos ( θ ) - 48 sin ( θ ) ) \times ∣ cos ( θ ) ∣}{48 cos ^{3} ( θ )}

Evaluate

y = 2 x^{2} - 16 x^{3} \times 3

Simplify

y = 2 x^{2} - 48 x^{3}

Move the expression to the left side

y - 2 x^{2} + 48 x^{3} = 0

To convert the equation to polar coordinates,substitute x for r cos (θ) and y for r sin (θ)

sin (θ) \times r - 2 (cos (θ) \times r)^{2} + 48 (cos (θ) \times r)^{3} = 0

Factor the expression

48 cos^{3} (θ) \times r^{3} - 2 cos^{2} (θ) \times r^{2} + sin (θ) \times r = 0

Factor the expression

r (48 cos^{3} (θ) \times r^{2} - 2 cos^{2} (θ) \times r + sin (θ)) = 0

When the product of factors equals 0,at least one factor is 0

r = 0 48 cos^{3} (θ) \times r^{2} - 2 cos^{2} (θ) \times r + sin (θ) = 0

Solution

More Steps

r = 0 r = \frac{cos ^{2} ( θ ) + cos ( θ ) ( cos ( θ ) - 48 sin ( θ ) ) \times ∣ cos ( θ ) ∣}{48 cos ^{3} ( θ )} r = \frac{cos ^{2} ( θ ) - cos ( θ ) ( cos ( θ ) - 48 sin ( θ ) ) \times ∣ cos ( θ ) ∣}{48 cos ^{3} ( θ )}

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