Hint
Factor 3 from the first two terms. Compare the horizontal level \(k\) with the vertex height.
Worked solution
(a)
\(g(x)=3(x^2-6x)+35=3[(x-3)^2-9]+35=3(x-3)^2+8\).
Answer: \(3(x-3)^2+8\).
(b)
Since \(3(x-3)^2\ge0\), the least value is 8, attained when the square is zero, at \(x=3\).
Answer: Minimum \(8\), at \(x=3\).
(c)
The equation is \((x-3)^2=(k-8)/3\). Two distinct real roots require a positive right side: \(k>8\). Taking both signs gives \(x=3\pm\sqrt{(k-8)/3}\). At k=8 the root is repeated.
Answer: \(k>8\); \(x=3\pm\sqrt{(k-8)/3}\).
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: Factor 3 from the quadratic and linear terms.
- 1 mark: Use the shift x−3 with the correct compensation.
- 1 mark: Obtain the constant 8.
Alternative methods: Coefficient comparison or expansion of a proposed square is valid.
Common error: Subtracting 9 rather than 27 from the original constant.
(b) · 2 marks
- 1 mark: State minimum 8 using nonnegativity of the square.
- 1 mark: State x=3.
Alternative methods: A derivative with a valid minimum justification is accepted.
Common error: Confusing the x-coordinate with the minimum value.
(c) · 3 marks
- 1 mark: Require (k−8)/3 to be strictly positive.
- 1 mark: State k>8, excluding equality.
- 1 mark: Give both exact roots with the square root covering (k−8)/3.
Alternative methods: The discriminant gives the same condition; the quadratic formula may give equivalent roots.
Common error: Including k=8 among the two-distinct-root cases.
Total: 8 marks.