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 ). transformation of graph dse exercise
The graph of ( y = \cos x ) is transformed to ( y = 3\cos(2x - \pi) + 1 ). Describe the sequence. Start with ( y = x^2 - 4 ) (vertex at (0,-4), roots at ±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 ). Stationary points occur when ( g'(x)=0 )