1. **State the problem:** We need to find the values of $m$ such that the line $y = mx - 6$ is tangent to the curve $y = x^2 - 4x + 3$. Also, find the coordinates of the tangent points. 2. **Set the line equal to the curve:** Since the line is tangent, it touches the curve at exactly one point. So, set $$mx - 6 = x^2 - 4x + 3$$ 3. **Rearrange to form a quadratic equation:** $$x^2 - 4x + 3 - mx + 6 = 0$$ $$x^2 - (4 + m)x + 9 = 0$$ 4. **Condition for tangency:** A quadratic has exactly one solution when its discriminant $D = 0$. The discriminant is $$D = b^2 - 4ac$$ Here, $a=1$, $b=-(4+m)$, $c=9$. 5. **Calculate discriminant:** $$D = (-(4+m))^2 - 4(1)(9) = (4+m)^2 - 36$$ 6. **Set discriminant to zero:** $$ (4+m)^2 - 36 = 0 $$ $$ (4+m)^2 = 36 $$ $$ 4+m = \pm 6 $$ 7. **Solve for $m$:** - If $4 + m = 6$, then $m = 2$ - If $4 + m = -6$, then $m = -10$ 8. **Find tangent points:** Substitute $m$ back into the quadratic to find $x$. For $m=2$: $$x^2 - (4+2)x + 9 = x^2 - 6x + 9 = 0$$ $$x^2 - 6x + 9 = 0$$ Discriminant is zero, so one root: $$x = \frac{6}{2} = 3$$ Find $y$ on the line: $$y = 2(3) - 6 = 6 - 6 = 0$$ Tangent point: $(3, 0)$ For $m = -10$: $$x^2 - (4 - 10)x + 9 = x^2 + 6x + 9 = 0$$ $$x^2 + 6x + 9 = 0$$ Discriminant zero, root: $$x = \frac{-6}{2} = -3$$ Find $y$ on the line: $$y = -10(-3) - 6 = 30 - 6 = 24$$ Tangent point: $(-3, 24)$ **Final answer:** - Possible values of $m$ are $2$ and $-10$. - Corresponding tangent points are $(3, 0)$ and $(-3, 24)$ respectively.