To find the turning points of the function \( y = x^3 + 3x^2 - x - 3 \), we first need to compute the first derivative and set it to zero. This will help us locate where the slope of the curve is zero (i.e., potential turning points).
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Finding the first derivative: \[ \frac{dy}{dx} = 3x^2 + 6x - 1 \]
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Setting the first derivative equal to zero: \[ 3x^2 + 6x - 1 = 0 \]
To solve this quadratic equation, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3, b = 6, c = -1 \).
Calculate the discriminant: \[ b^2 - 4ac = 6^2 - 4(3)(-1) = 36 + 12 = 48 \]
Now, plug in the values: \[ x = \frac{-6 \pm \sqrt{48}}{2 \cdot 3} = \frac{-6 \pm 4\sqrt{3}}{6} = \frac{-3 \pm \frac{2\sqrt{3}}{3}}{3} = \frac{-3 \pm \frac{2\sqrt{3}}{3}}{3} \]
Approximating the roots: \[ x_1 = -2 - \frac{2\sqrt{3}}{3} \approx -3.15 \quad \text{(rounded to -3)} \] \[ x_2 = -2 + \frac{2\sqrt{3}}{3} \approx -0.15 \quad \text{(rounded to -1)} \]
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Finding the corresponding y-values: Now substitute \( x_1 \) and \( x_2 \) back into the original function \( y(x) \) to find the corresponding y-values at the turning points:
For \( x = -2 \): \[ y(-2) = (-2)^3 + 3(-2)^2 - (-2) - 3 = -8 + 12 + 2 - 3 = 3 \quad \text{(point (-2, 3))} \]
For \( x = -1 \): \[ y(-1) = (-1)^3 + 3(-1)^2 - (-1) - 3 = -1 + 3 + 1 - 3 = 0 \quad \text{(point (-1, 0))} \]
For \( x = 0 \): \[ y(0) = (0)^3 + 3(0)^2 - (0) - 3 = -3 \quad \text{(point (0, -3))} \]
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List of turning points: The turning points of the function, rounded to the nearest whole number, are:
- \((-2, 3)\)
- \((-1, 0)\)
- \((0, -3)\)
So the correct list of turning points is: (–2, 3), (–1, 0), and (0, –3).