To find the zeros of the quadratic equation \(y = -3x^2 + 12x + 15\), we need to set \(y\) equal to zero and solve for \(x\):
\[ -3x^2 + 12x + 15 = 0 \]
We can solve this quadratic equation using the quadratic formula, which is:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
In our equation, the coefficients are:
- \(a = -3\)
- \(b = 12\)
- \(c = 15\)
Now, let's calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = 12^2 - 4(-3)(15) = 144 + 180 = 324 \]
Since the discriminant is positive, there will be two real and distinct solutions.
Now, let's find the roots using the quadratic formula:
\[ x = \frac{{-12 \pm \sqrt{324}}}{{2(-3)}} \]
Calculating \(\sqrt{324}\):
\(\sqrt{324} = 18\)
Now substitute back into the formula:
\[ x = \frac{{-12 \pm 18}}{{-6}} \]
This gives us two cases to solve:
-
Case 1: \[ x = \frac{{-12 + 18}}{{-6}} = \frac{6}{{-6}} = -1 \]
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Case 2: \[ x = \frac{{-12 - 18}}{{-6}} = \frac{-30}{{-6}} = 5 \]
Thus, the zeros of the equation are \(x = -1\) and \(x = 5\).
This means the zeros of \(y = -3x^2 + 12x + 15\) are:
\[ (-1, 0) \text{ and } (5, 0) \]
The correct response is:
\((−1,0)(5,0)\)
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