Home / IB Mathematics SL 5.4 Tangents and normal AI SL Paper 2 – Exam Style Questions

IB Mathematics SL 5.4 Tangents and normal AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

Consider the function defined by \(f(x)=x^2-10x\). The graph of \(f\) passes through the point \(A(4,-24)\).
(a) (i) Find the gradient of the tangent to the graph of \(f\) at the point \(A\).
     (ii) Hence, write down the gradient of the normal to the graph of \(f\) at point \(A\). [3]
(b) Write down the equation of the normal to the graph of \(f\) at point \(A\). [1]
The normal to the graph of \(f\) at point \(A\) intersects the graph of \(f\) again at a second point \(B\).
(c) Find the coordinates of \(B\). [3]
▶️ Answer/Explanation
Markscheme (with concise working)

(a)

\(f'(x)=2x-10 \Rightarrow f'(4)=2(4)-10=8-10=-2\).
Tangent gradient at \(A\): \(\boxed{-2}\).
Normal gradient is the negative reciprocal of \(-2\): \(\boxed{\tfrac{1}{2}}\). M1 A1 A1

(b)

Through \(A(4,-24)\) with slope \(\tfrac12\): point–slope form \[ y+24=\tfrac12(x-4)\ \Longrightarrow\ \boxed{\,y=\tfrac12x-26\,}. \] A1

(c)

Intersections of the normal with the parabola solve \[ x^2-10x \;=\; \tfrac12x-26 \ \Longrightarrow\ x^2-10.5x+26=0. \] One root is \(x=4\) (point \(A\)), so the other is \(x=10.5-4=6.5\).
\(y=f(6.5)=6.5^2-10(6.5)=42.25-65=-22.75\).
Hence \( \boxed{B\,(6.5,\,-22.75)} \) (accept \((6.5,-22.8)\) to 3 s.f.). M1 A1 A1
Total Marks: 7
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