Hint
Complete the square with a negative leading coefficient. For the restricted range, compare the vertex and both endpoints.
Worked solution
(a)
\(-2(x^2-6x)-11=-2[(x-3)^2-9]-11=-2(x-3)^2+7\). The nonpositive square term is largest at zero, giving (3,7).
Answer: \(h(x)=-2(x-3)^2+7\); maximum point \((3,7)\).
(b)
\(-2(x-3)^2+7\ge0\) gives \((x-3)^2\le7/2\). Thus \(-\sqrt{7/2}\le x-3\le\sqrt{7/2}\), giving the stated closed interval.
Answer: \(3-\frac{\sqrt{14}}2\le x\le3+\frac{\sqrt{14}}2\).
(c)
The vertex x=3 is included, so the maximum is 7. The endpoint values are h(0)=−11 and h(4)=5. The minimum is −11; continuity gives the full interval [−11,7].
Answer: \(-11\le h(x)\le7\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Obtain the shift x−3.
- 1 mark: Obtain the completed form with constant 7.
- 1 mark: State maximum point (3,7).
Alternative methods: Differentiation may locate the point but the requested completed form is still required.
Common error: Treating the vertex as a minimum despite the negative coefficient.
(b) · 3 marks
- 1 mark: Reduce correctly to (x−3)²≤7/2.
- 1 mark: Find both exact boundary values.
- 1 mark: Give the interval between them with inclusive endpoints.
Alternative methods: Equivalent endpoints 3±√(7/2), or a quadratic-formula and sign argument, are accepted.
Common error: Using the exterior intervals or strict endpoints.
(c) · 2 marks
- 1 mark: Compare the included vertex and both endpoints.
- 1 mark: Give the complete closed range [−11,7].
Alternative methods: Monotonicity on each side of x=3 plus endpoint values is accepted.
Common error: Using the unrestricted lower bound of negative infinity.
Total: 8 marks.