Tangent Values Ca51Ce

1. **State the problem:** We need to find the values of $k$ such that the x-axis ($y=0$) is tangent to the curve given by $$y = x^2 - (k+3)x + (3k + 4).$$ 2. **Understand the condition for tangency:** The x-axis is tangent to the curve if the quadratic equation $$x^2 - (k+3)x + (3k + 4) = 0$$ has exactly one solution. This means the discriminant must be zero. 3. **Write the discriminant formula:** For a quadratic equation $$ax^2 + bx + c = 0,$$ the discriminant $$\Delta$$ is given by $$\Delta = b^2 - 4ac.$$ 4. **Identify coefficients:** Here, $$a = 1,$$ $$b = -(k+3),$$ and $$c = 3k + 4.$$ 5. **Calculate the discriminant:** $$\Delta = (-(k+3))^2 - 4(1)(3k + 4) = (k+3)^2 - 4(3k + 4).$$ 6. **Set discriminant to zero for tangency:** $$ (k+3)^2 - 4(3k + 4) = 0.$$ 7. **Expand and simplify:** $$ (k+3)^2 = k^2 + 6k + 9,$$ so $$ k^2 + 6k + 9 - 12k - 16 = 0,$$ which simplifies to $$ k^2 - 6k - 7 = 0.$$ 8. **Solve the quadratic equation for $k$:** Using the quadratic formula: $$ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2} = \frac{6 \pm \sqrt{36 + 28}}{2} = \frac{6 \pm \sqrt{64}}{2}.$$ 9. **Calculate the roots:** $$ k = \frac{6 \pm 8}{2}.$$ So, - $$k = \frac{6 + 8}{2} = \frac{14}{2} = 7,$$ - $$k = \frac{6 - 8}{2} = \frac{-2}{2} = -1.$$ **Final answer:** The values of $k$ for which the x-axis is tangent to the curve are $$\boxed{7 \text{ and } -1}.$$