At which points does the graph of (f(x)=x^2-6x+8) cut the (x)-axis?

Correct answer: ((2,0)) and ((4,0))

The direct answer is A: (2,0) and (4,0). Points on the x-axis have y=0, so solve x²−6x+8=0. Factor the quadratic: x²−6x+8=(x−2)(x−4). Thus x=2 or x=4. With y=0, the intercepts are (2,0) and (4,0). Option A is correct. Option B uses negative roots, but substituting −2 or −4 does not make the expression zero. Option C gives points on the y-axis because their x-coordinate is 0; these are not x-intercepts. Option D uses 1 and 8, which are not the roots of the quadratic. A quick check is f(2)=4−12+8=0 and f(4)=16−24+8=0. Memory cue: for x-axis intersections, set y or f(x) equal to zero, then write coordinates with y=0.