Question

How do you graph $y=\sec \left( x-\dfrac{\pi }{4} \right)$?

Answer 1

to find the graph for the given equation, we have to use $y=a\sec \left( bx-c \right)+d$ for finding the amplitude, period, phase shift, vertical shift. After finding these values we have to find some points to plot on the graph. Now, based on the amplitude, period, vertical shift and phase shift and some more points we will draw a graph for the given equation.

Complete step by step answer:
From the given question we are given a graph for the equation $y=\sec \left( x-\dfrac{\pi }{4} \right)$.
To draw a graph for the equation let us consider the above equation as equation (1).
Let us consider
$y=\sec \left( x-\dfrac{\pi }{4} \right).........\left( 1 \right)$
To draw the graph for equation (1), we have to use the equation $y=a\sec \left( bx-c \right)+d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.
So let us consider the equation as equation (2).
$y=a\sec \left( bx-c \right)+d...........\left( 2 \right)$
By comparing equation (1) and (2), we get

& \Rightarrow a=1.............\left( 3 \right) \\\ & \Rightarrow b=1..............\left( 4 \right) \\\ & \Rightarrow c=\dfrac{\pi }{4}.............\left( 5 \right) \\\ & \Rightarrow d=0..............\left( 6 \right) \\\ \end{aligned}$$ Now let us consider the above equations as equation (3), equation (4), equation (5) and equation (6) respectively. Since the graph of the function $$\sec $$ does not have a maximum or minimum value, therefore we can say that amplitude will become none. Amplitude: None Now let us find the period using the formula $$\dfrac{2\pi }{\left| b \right|}$$. Now from the equation (4) let us find the period for equation (1) Therefore, Period of equation (1) is $$2\pi $$ Let us consider Period = $$2\pi ...............\left( 7 \right)$$ Let us find the phase shift using the formula $$\dfrac{c}{x}$$. Now from the equation (5) let us find the vertical shift for equation (1). Therefore, Phase shift of equation (1) is $$\dfrac{\pi }{4}$$. Let us consider Phase shift = $$\dfrac{\pi }{4}..............\left( 8 \right)$$. Let us find a vertical shift from the formula d. As we know from equation (6) we can say that vertical shift is equal to $$0$$. Therefore, let us consider Vertical shift = $$0.................\left( 9 \right)$$. Now we can say that graphs of equation (1) can be graphed using amplitude, period, phase shift, vertical shift, and the points. Selecting a few points to graph, $$x$$| $$\text{f(x)}$$ ---|--- $$\dfrac{\pi }{4}\text{ }$$| $$1$$ $$\dfrac{5\pi }{4}$$| $$-1$$ $$\dfrac{9\pi }{4}\text{ }$$| $$1$$ $$\dfrac{13\pi }{4}$$| $$-1$$ $$\dfrac{17\pi }{4}$$| $$1$$ Therefore from equation (7), (8), (9) and above points we get  **Note:** To find a graph for the equation $$y=\sec \left( x-\dfrac{\pi }{4} \right)$$, we should be aware of graphs concept in trigonometry. We should be careful while comparing the given equation with $$y=a\sec \left( bx-c \right)+d$$. Students should avoid calculation mistakes while doing this problem.