For what value(s) or range of values of \( k \) does \( y=-3 x^{2}-2 x+k \) have (i) no \( x \)-intercepts, (ii) one \( x \)-intercept, (iii) two \( x \)-intercepts?

To determine the values of \( k \) for which the quadratic equation \( y = -3x^2 - 2x + k \) has different types of \( x \)-intercepts, we can use the discriminant of the quadratic equation, given by \( D = b^2 - 4ac \), where \( a = -3 \), \( b = -2 \), and \( c = k \). (i) For no \( x \)-intercepts, the discriminant must be less than 0: \[ D < 0 \implies (-2)^2 - 4(-3)(k) < 0 \implies 4 + 12k < 0 \implies k < -\frac{1}{3} \] (ii) For one \( x \)-intercept, the discriminant must be equal to 0: \[ D = 0 \implies (-2)^2 - 4(-3)(k) = 0 \implies 4 + 12k = 0 \implies k = -\frac{1}{3} \] (iii) For two \( x \)-intercepts, the discriminant must be greater than 0: \[ D > 0 \implies (-2)^2 - 4(-3)(k) > 0 \implies 4 + 12k > 0 \implies k > -\frac{1}{3} \] In summary: - No \( x \)-intercepts: \( k < -\frac{1}{3} \) - One \( x \)-intercept: \( k = -\frac{1}{3} \) - Two \( x \)-intercepts: \( k > -\frac{1}{3} \)