\( f(x)=-(x-3)^{2}+25 \) 1) What are the zeros of the function? Write the smaller \( x \) first, and the larger \( x \) second. smaller \( x= \) larger \( x= \) 2) What is the vertex of the parabola? ( Show Calculator

To find the zeros of the function \( f(x)=-(x-3)^{2}+25 \), we set \( f(x) = 0 \): \[ -(x-3)^2 + 25 = 0 \] Solving this: \[ (x-3)^2 = 25 \] Taking the square root: \[ x-3 = 5 \quad \text{or} \quad x-3 = -5 \] This gives us: \[ x = 8 \quad \text{and} \quad x = -2 \] So, the smaller \( x \) is \( -2 \) and the larger \( x \) is \( 8 \). smaller \( x= -2 \) larger \( x= 8 \) Now for the vertex. The function can be expressed in vertex form as \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( h=3 \) and \( k=25 \), so the vertex of the parabola is: Vertex: \( (3, 25) \)