1. **State the problem:** Find values of $m$ for which the line $y=mx-6$ is tangent to the curve $y=x^2-4x+3$, and find the points of tangency. 2. **Set the line equal to the curve:** Since the line is tangent to the curve, they touch at exactly one point, so solve $$mx - 6 = x^2 - 4x + 3.$$ Rearranging this, $$x^2 - 4x - mx + 3 + 6 = 0$$ $$x^2 - (m+4)x + 9 = 0.$$ 3. **Condition for tangency:** The quadratic equation must have exactly one solution, so its discriminant $ riangle$ must be zero. $$ riangle = b^2 - 4ac = (-(m+4))^2 - 4(1)(9) = (m+4)^2 - 36 = 0.$$ 4. **Solve for $m$:** $$(m+4)^2 = 36$$ $$m+4 = \\pm 6.$$ Hence, - Case 1: $m+4=6 \Rightarrow m=2$ - Case 2: $m+4=-6 \Rightarrow m=-10$ 5. **Find points of tangency for each $m$:** **Case $m=2$:** Solve $$x^2 - (2+4)x + 9 = x^2 - 6x + 9 = 0.$$ This factors as $$(x-3)^2=0,$$ so $$x=3.$$ Calculate $y$: $$y=mx-6=2(3)-6=6-6=0.$$ Point is $(3,0)$. **Case $m=-10$:** Solve $$x^2 - (-10+4)x + 9 = x^2 + 6x + 9 = 0.$$ This factors as $$(x+3)^2=0,$$ so $$x=-3.$$ Calculate $y$: $$y = -10(-3) - 6 = 30 - 6 = 24.$$ Point is $(-3,24)$. **Final answers:** The possible values of $m$ are $2$ and $-10$. The points of tangency are $(3,0)$ and $(-3,24)$, respectively.