20N.1.AHL.TZ0.H_2

pestleMathematicsAAHLPaper 120N· sl-5-4-tangents-and-normalsource ↗

Find the equation of the tangent to the curve $y = e^{2 x} - 3 x$ at the point where $x = 0$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(x = 0 \Rightarrow) y = 1$ (A1)

appreciate the need to find $\frac{d y}{d x}$ (M1)

$(\frac{d y}{d x} =) 2 e^{2 x} - 3$ A1

$(x = 0 \Rightarrow) \frac{d y}{d x} = - 1$ A1

$\frac{y - 1}{x - 0} = - 1 (y = 1 - x)$ A1

[5 marks]

Examiners’ report

[N/A]