AQA Paper 3 2023 June — Question 3 1 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyEasy -1.8 This is a 1-mark multiple choice question testing only the basic definition that a normal is perpendicular to the tangent. Since f'(3)=0 means horizontal tangent, the normal must be vertical (x=3). No calculation required, just recall of a single concept.
Spec1.07m Tangents and normals: gradient and equations

A curve with equation \(y = f(x)\) passes through the point \((3, 7)\) Given that \(f'(3) = 0\) find the equation of the normal to the curve at \((3, 7)\) Circle your answer. [1 mark] \(y = \frac{7}{3}x\) \(y = 0\) \(x = 3\) \(x = 7\)

Question 3:
AnswerMarks Guidance
3Circles correct answer 2.2 a
Question 3 Total1
QMarking instructions AO
Question 3:
3 | Circles correct answer | 2.2 a | R 1 | x = 3
Question 3 Total | 1
Q | Marking instructions | AO | Marks | Typical solution
A curve with equation $y = f(x)$ passes through the point $(3, 7)$

Given that $f'(3) = 0$ find the equation of the normal to the curve at $(3, 7)$

Circle your answer.

[1 mark]

$y = \frac{7}{3}x$    $y = 0$    $x = 3$    $x = 7$

\hfill \mbox{\textit{AQA Paper 3 2023 Q3 [1]}}