To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \( y = -3x^2 + 12x + 15 \), the coefficients are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)
Now we can substitute these values into the quadratic formula:
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Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = 12^2 - 4 \cdot (-3) \cdot 15 = 144 + 180 = 324 \]
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Since the discriminant is positive, we know there will be two real zeros.
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Now substitute into the quadratic formula: \[ x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3} \]
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Simplify \(\sqrt{324}\): \[ \sqrt{324} = 18 \]
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Substitute this back into the formula: \[ x = \frac{-12 \pm 18}{-6} \]
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Calculate the two values for \(x\):
For the positive case: \[ x_1 = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
For the negative case: \[ x_2 = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
Thus, the zeros of the function \( y = -3x^2 + 12x + 15 \) are \( (-1, 0) \) and \( (5, 0) \).
The correct answer is: b (-1,0)(5,0).
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