To find the turning points of the function \( y = 4x^4 - 8x^2 + 4 \), we need to calculate its derivative and set it equal to zero to find the critical points.
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Find the derivative: \[ y' = \frac{d}{dx}(4x^4 - 8x^2 + 4) = 16x^3 - 16x \]
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Set the derivative equal to zero: \[ 16x^3 - 16x = 0 \] Factor out the common term: \[ 16x(x^2 - 1) = 0 \] This gives us: \[ 16x(x-1)(x+1) = 0 \]
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Solve for \( x \): \[ x = 0, \quad x = 1, \quad x = -1 \]
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Find the corresponding \( y \)-values at these \( x \)-values:
- For \( x = 0 \): \[ y(0) = 4(0)^4 - 8(0)^2 + 4 = 4 \]
- For \( x = 1 \): \[ y(1) = 4(1)^4 - 8(1)^2 + 4 = 4 - 8 + 4 = 0 \]
- For \( x = -1 \): \[ y(-1) = 4(-1)^4 - 8(-1)^2 + 4 = 4 - 8 + 4 = 0 \]
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Summarize the turning points: From the calculations above, the turning points are:
- \( (0, 4) \)
- \( (1, 0) \)
- \( (-1, 0) \)
Thus, the turning points of the graph are:
(–1, 0), (0, 4), and (1, 0)
The correct option is: (–1, 0), (0, 4), and (1, 0).