To locate the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we need to set the equation to zero and solve for \( x \):
\[ -3x^2 + 12x + 15 = 0 \]
Since it's a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = -3 \), \( b = 12 \), and \( c = 15 \), we can either factor it, complete the square, or use the quadratic formula. Let's use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
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Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 = 12^2 = 144 \] \[ 4ac = 4 \cdot (-3) \cdot 15 = -180 \] \[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
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Apply the quadratic formula: \[ x = \frac{-12 \pm \sqrt{324}}{2 \cdot -3} \] \[ \sqrt{324} = 18 \]
So the equation becomes: \[ x = \frac{-12 \pm 18}{-6} \]
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Calculate the two possible values for \( x \):
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For the positive root: \[ x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
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For the negative root: \[ x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
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Thus, the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \) are:
\[ x = -1 \quad \text{and} \quad x = 5 \]