To locate the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we need to set the equation equal to zero and solve for \( x \):
\[ -3x^2 + 12x + 15 = 0 \]
We can first multiply the entire equation by -1 (to make calculations simpler):
\[ 3x^2 - 12x - 15 = 0 \]
Next, we can use the quadratic formula to find the zeros of the equation, where \( a = 3 \), \( b = -12 \), and \( c = -15 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = (-12)^2 - 4(3)(-15) = 144 + 180 = 324 \]
Now apply the quadratic formula:
\[ x = \frac{12 \pm \sqrt{324}}{2 \times 3} \]
Calculating \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Now, substituting back into the formula:
\[ x = \frac{12 \pm 18}{6} \]
This gives us two potential solutions:
- \( x = \frac{12 + 18}{6} = \frac{30}{6} = 5 \)
- \( x = \frac{12 - 18}{6} = \frac{-6}{6} = -1 \)
So the zeros of the function are \( x = 5 \) and \( x = -1 \).
The respective points where these zeros occur on the graph are:
\((-1, 0)\) and \((5, 0)\).
Therefore, the correct answer is:
(-1, 0) and (5, 0).
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