Transformation Of Graph Dse Exercise May 2026

Question 5 (M2 Type):
Given ( f(x) = x^3 - 3x ). Let ( g(x) = 2f(1-x) + 1 ). Find the x-coordinates of stationary points of ( g ).

Method:
Stationary points occur when ( g'(x)=0 ).
( g(x) = 2f(1-x) + 1 )
( g'(x) = 2 \cdot f'(1-x) \cdot (-1) = -2 f'(1-x) )
Set ( g'(x)=0 \implies f'(1-x)=0 ).

Now ( f'(x)=3x^2-3 = 3(x^2-1) ).
So ( f'(1-x)=0 \implies (1-x)^2 - 1 =0 \implies (1-x)^2=1 )
( \implies 1-x = \pm 1 \implies x=0 ) or ( x=2 ).

Thus stationary points at ( x=0, 2 ).

Translation involves shifting the graph horizontally or vertically without altering its shape or orientation.

A. Vertical Translation

B. Horizontal Translation

DSE Exam Tip: Remember "Inside changes opposite, outside changes direct."

DSE often combines ( y = |f(x)| ) or ( y = f(|x|) ) with other transformations.

Question 7:
Sketch ( y = |x^2 - 4| - 1 ). How many x-intercepts? transformation of graph dse exercise

Step 1: Start with ( y = x^2 - 4 ) (vertex at (0,-4), roots at ±2).
Step 2: Apply modulus: ( y = |x^2 - 4| ) – reflect negative part above x-axis.
Step 3: Subtract 1: shift graph down by 1.

Analysis for intercepts:
Solve ( |x^2 - 4| - 1 = 0 \implies |x^2 - 4| = 1 )
Two cases:

The graph of ( y = h(x) ) is transformed by a reflection in the x-axis, followed by a horizontal translation of 3 units left, giving ( y = \frac1x+2 ). Find ( h(x) ). Question 5 (M2 Type): Given ( f(x) = x^3 - 3x )