Hint
Evaluate the ordinate from the original curve. Differentiate and evaluate at the stated point.
Worked solution
(a)
\(2 x - 6\) gives gradient \(2\) at \(\left( 4, \ 15\right)\). Substitute into the point-gradient form.
Answer: \(y=2 x + 7\)
(b)
The normal gradient is \(\displaystyle - \frac{1}{2}\). The line must pass through \(\left( 4, \ 15\right)\).
Answer: \(\displaystyle y=17 - \frac{x}{2}\)
(c)
Use y=mx+b at the known point.
\(7\)
Answer: \(7\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Evaluate the derivative and ordinate at the specified abscissa and use point-gradient form.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A normal gradient has the opposite sign.
(b) · 2 marks
- 1 mark: Use the negative reciprocal of the nonzero tangent gradient through the curve point.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: The tangent intercept is not usually f(a).
(c) · 2 marks
- 1 mark: Use y=mx+b at the known point.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A normal gradient has the opposite sign.
Total: 6 marks.