(a) 求曲线 $y=3x^2-12x+17$ 与直线 $y=6x+2$ 的交点坐标。

(b) 由此，解不等式 $6x+2\ge3x^2-12x+17$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

交点的纵坐标相同，所以令两式相等。先求横坐标，再用直线回代。

Equating the two expressions gives

Thus $x=1$ or $x=5$. Substituting into $y=6x+2$ gives the points $(1,8)$ and $(5,32)$.

用 (a) 的两个交点作为分界。用曲线高度减去直线高度，所得二次式应小于或等于零。

The inequality is equivalent to $3(x-1)(x-5)\le0$. This holds between the roots, including the endpoints, so $1\le x\le5$.