To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can set the equation to zero and solve for \( x \):
\[ -3x^2 + 12x + 15 = 0 \]
First, let's simplify the equation by multiplying through by -1 to make the leading coefficient positive:
\[ 3x^2 - 12x - 15 = 0 \]
Next, we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \( a = 3 \), \( b = -12 \), and \( c = -15 \). Plugging these values into the formula:
\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(-15)}}{2(3)} \]
Calculating step by step:
- \( -(-12) = 12 \)
- \( (-12)^2 = 144 \)
- \( 4(3)(-15) = -180 \)
- Hence, \( b^2 - 4ac = 144 + 180 = 324 \)
Now, substituting these into the formula:
\[ x = \frac{12 \pm \sqrt{324}}{6} \]
Calculate \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Now substituting back into the equation:
\[ x = \frac{12 \pm 18}{6} \]
This gives us two solutions:
- \( x = \frac{12 + 18}{6} = \frac{30}{6} = 5 \)
- \( x = \frac{12 - 18}{6} = \frac{-6}{6} = -1 \)
Thus, the zeros of the function are \( x = 5 \) and \( x = -1 \).
Hence, the coordinates of the zeros are:
\[ (-1, 0) \text{ and } (5, 0) \]
So the correct response is:
\[ \text{(−1,0)(5,0) } \]
Which reads as \( \left( -1, 0 \right) \quad \left( 5, 0 \right) \)