For $f(x)=\dfrac{2x-3}{x^2}$ , with $f'(x)=\dfrac{-2x+6}{x^3}$ , find the equation of the tangent to the curve at $x=2$ .

Model answer

4y=x-1

Also accepted: y=x/4-1/4, 4y = x - 1

Explanation:

f(2)=\tfrac{4-3}4=\tfrac14

;

f'(2)=\tfrac{-4+6}8=\tfrac14

. Tangent:

y-\tfrac14=\tfrac14(x-2)\Rightarrow4y=x-1

Derived from ZIMSEC Pure Mathematics Paper 2, November 2025, Q3

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