Transformation Of Graph Dse Exercise <Genuine>

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

Thus stationary points at ( x=0, 2 ). Trig graphs test horizontal scaling (period change) and vertical scaling (amplitude) most intensely.

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. transformation of graph dse exercise

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.

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 ). The graph of ( y = \cos x

The graph of ( y = f(x) ) is translated 3 units right and then reflected in the y-axis to become ( y = \sqrt4 - x^2 ). Find ( f(x) ).

Sketch ( y = |x^2 - 4| - 1 ). How many x-intercepts? Trig graphs test horizontal scaling (period change) and

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 ).