\( f(r)=-(r-12)(r+3) \) 1) What are the zeros of the function? Write the smaller \( r \) first, and the larger \( r \) second. smaller \( r= \) larger \( r= \) 2) What is the vertex of the parabola? 2.

To find the zeros of the function \( f(r)=-(r-12)(r+3) \), we set the equation to zero: \[ -(r-12)(r+3) = 0 \] This means either \( r-12 = 0 \) or \( r+3 = 0 \). 1) Solving these gives us: - \( r-12=0 \Rightarrow r=12 \) - \( r+3=0 \Rightarrow r=-3 \) Thus, the zeros of the function are: smaller \( r= -3 \) larger \( r= 12 \) To find the vertex of the parabola described by \( f(r) \), we can use the formula for the vertex \( r = -\frac{b}{2a} \) when the function is in standard form \( ar^2 + br + c \). First, let's expand \( f(r) \): \[ f(r) = -r^2 + 9r - 36 \] This gives \( a = -1 \) and \( b = 9 \). The vertex occurs at: \[ r = -\frac{9}{2(-1)} = \frac{9}{2} = 4.5 \] To find the y-coordinate of the vertex, plug \( r = 4.5 \) back into the function: \[ f(4.5) = -(4.5-12)(4.5+3) = -(-7.5)(7.5) = -(-56.25) = 56.25 \] So the vertex is at: \( (4.5, 56.25) \).