If the image of function y = KX's square-6x + 3 intersects with X axis, then the value range of K is () A.K < 3b.k < 3 If the image of function y = KX square-6x + 3 intersects with X axis, then the value range of K is () A.K < 3 b.k < 3 and K ≠ 0 C.K ≤ 3 D.K ≤ 3 and K ≠ 0

If the image of function y = KX's square-6x + 3 intersects with X axis, then the value range of K is () A.K < 3b.k < 3 If the image of function y = KX square-6x + 3 intersects with X axis, then the value range of K is () A.K < 3 b.k < 3 and K ≠ 0 C.K ≤ 3 D.K ≤ 3 and K ≠ 0

∵ the image of quadratic function y = kx2-6x + 3 has intersection with X axis,
The equation kx2-6x + 3 = 0 (K ≠ 0) has real roots,
That is, △ = 36-12k ≥ 0, K ≤ 3, because it is a quadratic function, so K ≠ 0, then the value range of K is k ≤ 3 and K ≠ 0
So D
Do not understand can ask, help please adopt, thank you!