To locate the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we need to find the values of \( x \) where \( y = 0 \). This means we need to solve the equation:
\[ -3x^2 + 12x + 15 = 0 \]
We can use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, the coefficients \( a \), \( b \), and \( c \) are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)
Now, we will plug these values into the quadratic formula.
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 12^2 = 144 \] \[ 4ac = 4(-3)(15) = -180 \] \[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
- Now plug the values into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{324}}{2(-3)} \]
- Calculate \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
- Substitute back into the formula:
\[ x = \frac{-12 \pm 18}{-6} \]
- Calculate the two possible values for \( x \):
First solution:
\[ x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
Second solution:
\[ x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
Thus, the zeros of the function \( y = -3x^2 + 12x + 15 \) are:
\[ x = -1 \quad \text{and} \quad x = 5 \]
Summary:
The zeros of the function are \( x = -1 \) and \( x = 5 \).