Question

determined the zeros of the quadric function y=-x^2 + 15x - 36

Answer 1

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 \( x = 12 \).