Hint

First recover the family of curves. The supplied point selects the constant; the gradient still comes from the given derivative.

Worked solution

(a)

Integrating \(\displaystyle 3x^{2}-4x+2\) term by term gives \(\displaystyle y=x^{3}-2x^{2}+2x+C\).

Answer: \(y=x^3-2x^2+2x+C\).

(b)

At \(\displaystyle (1,5), 5=1-2+2+C=1+C\), so \(\displaystyle C=4\). The curve is \(\displaystyle y=x^{3}-2x^{2}+2x+4\), whose derivative is the original gradient.

Answer: \(y=x^3-2x^2+2x+4\).

(c)

The point has ordinate \(\displaystyle 8-8+4+4=8\). The given derivative at \(\displaystyle x=2\) is \(\displaystyle 12-8+2=6\). Thus \(\displaystyle y-8=6(x-2)\), giving \(\displaystyle y=6x-4\).

Answer: \(y=6x-4\).

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: Integrate the cubic-producing term as \(\displaystyle x^{3}\).
• 1 mark: Integrate the remaining terms as \(\displaystyle -2x^{2}+2x\).
• 1 mark: Retain the arbitrary constant C.

Alternative methods: Any equivalent antiderivative family is accepted.

Common error: Writing the derivative itself as the curve equation.

(b) · 2 marks

• 1 mark: Use the supplied point to obtain \(\displaystyle C=4\).
• 1 mark: Give the complete curve equation.

Alternative methods: An equivalent curve equation satisfying both conditions is accepted.

Common error: Substituting the point ordinate into \(\displaystyle \frac{dy}{dx}\) to find C.

(c) · 3 marks

• 1 mark: Find the point (2,8) on the recovered curve.
• 1 mark: Find tangent slope 6.
• 1 mark: Give the correct tangent line through that point.

Alternative methods: The exact point–slope equation is accepted.

Common error: Using the initial point (1,5) for a tangent requested at \(\displaystyle x=2\).

Total: 8 marks.