Question 7 Consider the Quadratic function \( f(x)=-8(x-5)(x-3) \). Its vertex is:

To find the vertex of the quadratic function \( f(x) = -8(x-5)(x-3) \), we can first rewrite it in standard form. The vertex of a quadratic function in the form \( f(x) = a(x-h)^2 + k \) is given by the point \( (h, k) \). 1. **Expand the function**: \[ f(x) = -8(x-5)(x-3) \] First, we expand \( (x-5)(x-3) \): \[ (x-5)(x-3) = x^2 - 3x - 5x + 15 = x^2 - 8x + 15 \] Now, substitute this back into the function: \[ f(x) = -8(x^2 - 8x + 15) = -8x^2 + 64x - 120 \] 2. **Find the vertex**: The vertex \( x \)-coordinate can be found using the formula \( x = -\frac{b}{2a} \), where \( a = -8 \) and \( b = 64 \): \[ x = -\frac{64}{2 \cdot -8} = -\frac{64}{-16} = 4 \] 3. **Calculate the \( y \)-coordinate**: Now, substitute \( x = 4 \) back into the function to find the \( y \)-coordinate: \[ f(4) = -8(4-5)(4-3) = -8(-1)(1) = 8 \] Thus, the vertex of the quadratic function \( f(x) = -8(x-5)(x-3) \) is at the point \( (4, 8) \). **Final Answer**: The vertex is \( (4, 8) \).