Transformation Of Graph Dse Exercise Official

Introduction: Why Graph Transformations Matter in DSE In the Hong Kong DSE Mathematics examination, the ability to manipulate and interpret graphs is not merely a mechanistic skill—it is a visual language. Questions involving transformation of graphs appear consistently across Papers 1 (Conventional) and 2 (MCQ), as well as in the M2 Calculus paper.

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

A and D are equivalent and correct. Reflection first: ( y = -\sin x ), then +2. Exercise Set 2: Finding the Original Graph (Reverse Transformation) DSE often asks: Given the image graph, find the pre-image function. Introduction: Why Graph Transformations Matter in DSE In

The graph of ( y = \cos x ) is transformed to ( y = 3\cos(2x - \pi) + 1 ). Describe the sequence. How many x-intercepts

The graph of ( y = 2^x ) is reflected in the line ( y = x ), then stretched vertically by factor 3, then translated 2 units down. Find the equation of the resulting curve. Answer: Reflection in ( y=x ) gives inverse: ( y = \log_2 x ). Then vertical stretch ×3: ( y = 3 \log_2 x ). Then down 2: ( y = 3 \log_2 x - 2 ).

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 ). Trig graphs test horizontal scaling (period change) and vertical scaling (amplitude) most intensely.

Introduction: Why Graph Transformations Matter in DSE In the Hong Kong DSE Mathematics examination, the ability to manipulate and interpret graphs is not merely a mechanistic skill—it is a visual language. Questions involving transformation of graphs appear consistently across Papers 1 (Conventional) and 2 (MCQ), as well as in the M2 Calculus paper.

Sketch ( y = |x^2 - 4| - 1 ). How many x-intercepts?

A and D are equivalent and correct. Reflection first: ( y = -\sin x ), then +2. Exercise Set 2: Finding the Original Graph (Reverse Transformation) DSE often asks: Given the image graph, find the pre-image function.

The graph of ( y = \cos x ) is transformed to ( y = 3\cos(2x - \pi) + 1 ). Describe the sequence.

The graph of ( y = 2^x ) is reflected in the line ( y = x ), then stretched vertically by factor 3, then translated 2 units down. Find the equation of the resulting curve. Answer: Reflection in ( y=x ) gives inverse: ( y = \log_2 x ). Then vertical stretch ×3: ( y = 3 \log_2 x ). Then down 2: ( y = 3 \log_2 x - 2 ).

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 ). Trig graphs test horizontal scaling (period change) and vertical scaling (amplitude) most intensely.