2. Calculate the value(s) of \( k \) for which \( -x^{2}+6 x-k \) has real roots. Answer: \( k \) Select \( v \)

To determine the values of \( k \) for which the quadratic equation \( -x^{2}+6x-k \) has real roots, we can utilize the discriminant from the quadratic formula, which states that for an equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( D = b^2 - 4ac \) must be greater than or equal to zero. In this case, \( a = -1 \), \( b = 6 \), and \( c = -k \). The discriminant is given by: \[ D = 6^2 - 4(-1)(-k) = 36 - 4k \] To have real roots, set the discriminant \( D \geq 0 \): \[ 36 - 4k \geq 0 \\ 4k \leq 36 \\ k \leq 9 \] Therefore, \( k \) must be less than or equal to 9 for the quadratic to have real roots. To ensure that the quadratic is usable in practical situations, always remember that the vertex can also determine the nature of the roots based on whether it lies above or below the x-axis. The vertex of this equation is pivotal: it gives a critical point that can help visualize whether roots exist in given intervals based on where it sits in relation to zero. So, check vertex location while using your discriminant findings!