Graph y=sec(2x+(3pi)/4)+1

Problem

y=sec(2*x+(3*π)/4)+1

Solution

1. Identify the parent function and its properties. The function is a transformation of y=sec(u) which has vertical asymptotes where cos(u)=0

2. Determine the amplitude and vertical shift. The coefficient of the secant term is 1 and the constant d=1 shifts the graph upward by 1 unit. The midline is y=1

3. Calculate the period. The coefficient of x is b=2 The period P is calculated as:

P=(2*π)/|b|

P=(2*π)/2=π

4. Find the phase shift. Factor out the coefficient of x inside the argument:

2*x+(3*π)/4=2*(x+(3*π)/8)

The phase shift is (3*π)/8 units to the left.

5. Locate the vertical asymptotes. Asymptotes occur where the inner argument equals π/2+n*π

2*x+(3*π)/4=π/2

2*x=−π/4

x=−π/8

Adding half a period (π/2 gives the next asymptote at x=(3*π)/8

6. Identify key points. The local minima and maxima of the secant function correspond to the peaks and valleys of the reciprocal cosine function. A local minimum occurs midway between x=−π/8 and x=(3*π)/8 at x=π/8

y=sec(2*(π/8)+(3*π)/4)+1

y=sec(π)+1=−1+1=0

The point is (π/8,0)

Final Answer

y=sec(2*x+(3*π)/4)+1

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