Below is a graph of the function y = sec ⁡ ( b x − π ) y=\sec\left(bx-\pi\right) y = sec ( b x − π ) . Determine the value of b .

Let's give this problem a try. Here we're told below is a graph of the function y is equal to the secant of bx minus pi determine the value of b. Now to solve this problem, when dealing with these kinds of trigonometric graphs, what I like to do is start with a trig function I'm more familiar with, and then work off of that one. Now what I can see from this is that we have a secant. And I know that secant is a reciprocal identity related to the cosine. So I'm first going to find the graph y is equal to the cosine of bx minus pi. Now rather than trying to figure out what this graph exactly looks like by dealing with this kind of phase shift, and then moving the graph around. What I'm going to do is recognize that for this cosine function, it's going to have valleys. At all of these places we have these kind of downward curves, it's going to have peaks at all these places we have upward curves. So the cosine function is gonna start at negative 1, and it's going to go up and reach a peak at pi over 2. Then Then it's gonna come down here and reach a valley at pi. We're going to peak again at 3 pi over 2. We're going to valley down here at 2 pi, and then we're going to reach a peak again at 5 pi over 2. So this is what the cosine function is going to look like when we graph it. Now what I can do is I can recognize that this cosine function starts at a low value, and then it goes up, and then it goes back down, and it completes one full period when we get to pi on the x axis. So I can see that the period of this cosine function is pi. Now why do I need the period? Well that's because I know that the period of a cosine function is 2pi over b. So what I can do is use this equation to find 'b' by recognizing that the period is pi. So I'm going to set the period equal to pi, so we'll have that pi is equal to 2pi over b, and then all I need to do from here is solve for b. Now I'm gonna do this by first multiplying b on both sides of the equation. That'll get the b's to cancel on the right, giving me pi times b is equal to 2 pi. Now from here what I can do is divide both sides of this equation by pi. And doing this will allow me to cancel the pi's on the left side and on the right side. So all I'm going to have left over is 'b' is equal to 2. So this right here is the solution for 'b', and the answer to this problem. So hope you found this video helpful, thanks for watching.

Below is a graph of the function y = sec ⁡ ( b x − π ) y=\sec\left(bx-\pi\right) y = sec ( b x − π ) . Determine the value of b .

b = π 2 b=\frac{\pi}{2} b = 2 π

10. Graphing Trigonometric Functions

Graphs of Secant and Cosecant Functions

Precalculus

10. Graphing Trigonometric Functions

Graphs of Secant and Cosecant Functions

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Multiple Choice

Below is a graph of the function $y=\sec\left(bx-\pi\right)$ . Determine the value of b.

$b=2$

$b=4$

$b=\frac{\pi}{2}$

$b=\pi$

Verified step by step guidance

Identify the period of the secant function from the graph. The period is the distance between two consecutive vertical asymptotes. In this graph, the vertical asymptotes occur at x = π/2, x = 3π/2, and x = 5π/2, indicating a period of π.

Recall the general form of the secant function y = sec(bx - π). The period of the secant function is given by the formula 2π/b.

Set the period from the graph equal to the formula for the period: π = 2π/b.

Solve the equation π = 2π/b for b. Divide both sides by π to isolate b, resulting in 1 = 2/b.

Multiply both sides by b and then divide by 2 to solve for b, yielding b = 2.

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Struggling with Precalculus?

Below is a graph of the function y=sec⁡(bx−π)y=\sec\left(bx-\pi\right)y=sec(bx−π). Determine the value of b.

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Below is a graph of the function $y=\sec\left(bx-\pi\right)$ . Determine the value of b.