To find the zeros of the quadratic function \( y = -x^2 + 15x - 36 \), we can set the function equal to zero and solve for \( x \):
\[ -x^2 + 15x - 36 = 0 \]
To make it easier to solve, we can multiply the entire equation by -1:
\[ x^2 - 15x + 36 = 0 \]
Next, we can factor the quadratic:
We need two numbers that multiply to \( 36 \) (the constant term) and add up to \(-15\) (the coefficient of \( x \)). The numbers \(-3\) and \(-12\) satisfy this, so we can factor the quadratic as follows:
\[ (x - 3)(x - 12) = 0 \]
Setting each factor equal to zero gives us the zeros:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \]
Thus, the zeros of the quadratic function \( y = -x^2 + 15x - 36 \) are \( x = 3 \) and \( x = 12 \).