1. Problem (a): Express $2x^2 - 12x + 3$ in the form $a(x+b)^2 + c$ where $a,b,c$ are integers. 2. Use the method of completing the square. 3. Start with the expression: $$2x^2 - 12x + 3$$ 4. Factor out 2 from the first two terms: $$2(x^2 - 6x) + 3$$ 5. Complete the square inside the parentheses: - Take half of $-6$, which is $-3$, and square it: $(-3)^2 = 9$ 6. Add and subtract 9 inside the parentheses: $$2(x^2 - 6x + 9 - 9) + 3$$ 7. Group the perfect square trinomial and simplify: $$2((x - 3)^2 - 9) + 3$$ 8. Distribute 2: $$2(x - 3)^2 - 18 + 3$$ 9. Simplify constants: $$2(x - 3)^2 - 15$$ 10. So, $a=2$, $b=-3$, $c=-15$. --- 11. Problem (b): The curve $C$ has equation $$y = 2(x + 4)^2 - 12(x + 4) + 3$$ 12. Find the minimum point $M$. 13. Let $t = x + 4$, rewrite $y$: $$y = 2t^2 - 12t + 3$$ 14. Complete the square for $y$ in terms of $t$: $$2(t^2 - 6t) + 3$$ 15. Half of $-6$ is $-3$, square is 9: $$2(t^2 - 6t + 9 - 9) + 3 = 2((t - 3)^2 - 9) + 3$$ 16. Simplify: $$2(t - 3)^2 - 18 + 3 = 2(t - 3)^2 - 15$$ 17. Minimum occurs when $(t - 3)^2 = 0$, so $t = 3$. 18. Recall $t = x + 4$, so $x + 4 = 3 ightarrow x = -1$. 19. Find $y$ at $x = -1$: $$y = 2(3)^2 - 12(3) + 3 = 2(9) - 36 + 3 = 18 - 36 + 3 = -15$$ 20. Coordinates of $M$ are $(-1, -15)$. --- 21. Problem (11): Solve inequality $$2y^2 - 7y - 30 \leq 0$$ 22. Factor the quadratic: Find factors of $2 \times (-30) = -60$ that sum to $-7$: $-12$ and $5$. 23. Rewrite: $$2y^2 - 12y + 5y - 30 \leq 0$$ 24. Group terms: $$(2y^2 - 12y) + (5y - 30) \leq 0$$ 25. Factor each group: $$2y(y - 6) + 5(y - 6) \leq 0$$ 26. Factor out $(y - 6)$: $$(2y + 5)(y - 6) \leq 0$$ 27. Find roots: $$2y + 5 = 0 \Rightarrow y = -\frac{5}{2}$$ $$y - 6 = 0 \Rightarrow y = 6$$ 28. Test intervals: - For $y < -\frac{5}{2}$, pick $y = -3$: $(2(-3)+5)(-3-6) = (-1)(-9) = 9 > 0$ - For $-\frac{5}{2} < y < 6$, pick $y=0$: $(5)(-6) = -30 < 0$ - For $y > 6$, pick $y=7$: $(19)(1) = 19 > 0$ 29. Inequality holds where product is $\leq 0$, so: $$-\frac{5}{2} \leq y \leq 6$$ Final answers: (a) $2(x - 3)^2 - 15$ (b) $M = (-1, -15)$ (11) $-\frac{5}{2} \leq y \leq 6$