Transformation Of Graph Dse Exercise _hot_ Jun 2026

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 ).

Let the original function be ( f(x) ).

is translated 2 units to the left, then compressed vertically by a factor of 0.5, and finally reflected across the x-axis, find the equation of the new graph Translate left by 2: Compress vertically by 0.5: Reflect across x-axis: Result: transformation of graph dse exercise

The graph of $y = f(x)$ undergoes the following transformations in order: Now ( f'(x)=3x^2-3 = 3(x^2-1) )

Thus ( f(x) = x^2 - 4x + 5 ).