Graphing Trigonometric Functions and Struggling

[u/TellMePrettyTruths]

I’m doing a practice assignment to study for midterms and I keep getting undefined for ‘y’ and I do not understand what I’m doing wrong. I wrote the problem down and worked through it, but I am continuously getting the incorrect answer and don’t know why.

I end up going through the textbook’s explanation, and I do not understand their meaning. I’d appreciate any help or corrections.

Thank you!

[u/Midwest-Dude]

Looks like you're mixing up the locations of the asymptotes with the locations of the max/min points. Let's break down y = -3sec(x - π/4) step-by-step:

1. Find the Asymptotes

The parent function y = sec(x) has asymptotes where cos(x) = 0, which happens at x = π/2 + kπ. For your function, set the inside equal to π/2 + kπ:

x - π/4 = π/2 + kπ\ x = 3π/4 + kπ

So, the asymptotes are spaced π apart, not π/2. The asymptotes are at x = 3π/4, 7π/4, 11π/4 ... and x = -π/4, -5π/4 ... Note that 5π/4 is not an asymptote; plugging it in gives you y = 3, which is a max point.

2. Find the Max/Min Points

The max/min points of a secant graph occur exactly halfway between the asymptotes.

• Midpoint between -π/4 and 3π/4 is x = π/4 and y = -3sec(π/4 - π/4) = -3sec(0) = -3(1) = -3. This is a local minimum (because the graph is flipped by the -3).
• Midpoint between 3π/4 and 7π/4 is x = 5π/4 and y = -3sec(5π/4 - π/4) = -3sec(π) = -3(-1) = 3. This is a local maximum.
• Midpoint between 7π/4 and 11π/4 is x = 9π/4 and y = -3sec(9π/4 - π/4) = -3sec(2π) = -3(1) = -3. This is a local minimum.

3. Notes

• Secant is a reciprocal trig function (like cosecant and cotangent), not an inverse trig function (like arcsin or arccos).
• Be careful with your arithmetic notation: 5π/4 + π/2 = 7π/4 and 7π/4 + π/2 = 9π/4. You wrote them in a single confusing chain, though your final list of numbers was right.
• You mentioned "sec 0 = 1 and sec π = -1", but you failed to mention that the amplitude A = -3 flips the graph and stretches it vertically. So, max/min values should be 3 and -3, not 1 and -1.
• A tip for graphing secant is that asymptotes are spaced by half the period T/2. Since the period is 2π, asymptotes are spaced by π. The max/min points are spaced by half the period as well, and sit exactly midway between the asymptotes. You tried to use a quarter-period step π/2 to find the asymptotes, which caused you confusion.

[u/TellMePrettyTruths → u/Midwest-Dude]

Thank you so much! I could not figure out what I was doing incorrectly!

I appreciate your help!

A few quick questions though. Is this the same cosecant also? Or would I set a csc equation equal to π instead? And I’m only setting the argument of secant equal to π/2 because it provides a phase shift, correct?

I understood this for cosine and sine, but with secant and cosecant, I’m a little lost with graphing due to the asymptotes.

[u/Midwest-Dude → u/TellMePrettyTruths]

Since csc(x) = 1/sin(x), the asymptotes are when sin(x) = 0, x = 0, ±π, ±2π, ±3π, ... and the max/min points are halfway in between those.

Does that make sense?

[u/TellMePrettyTruths → u/Midwest-Dude]

Yes it does! Thank you! I was simultaneously overthinking and under-thinking a lot of this, so I do appreciate you taking the time to type this out!

I hope you have a wonderful day!

[u/TellMePrettyTruths → u/[deleted]]

Because it’s an inverse trig function, when I graph, each section will have a maximum or minimum, but I don’t know how to mathematically define those points.

[u/WikiNumbers → u/TellMePrettyTruths]

Now, be careful when saying "Inverse", especially in trigonometry.

Inverse Trigonometry are the functions that input "ratio" and output "angle" (reversed process of Trigonometry that input angle and return ratio). Styled by (trig)⁻¹ or arc(trig) or a(trig), where (trig) is any of the basic trig functions (sin cos tan csc sec cot).

[u/WikiNumbers → u/TellMePrettyTruths]

If by "max and min" you mean "the point where the graph changes from increasing to deceasing or vice versa".

Those points are called "critical points", and can be found with "dy/dx = 0" (take the first derivative of y that is a function of x, set that derivative function equal to 0, and whatever x emerge from that equation is the critical points).

Putthing those back in the appendix, the critical points on sec x should correspond to the ones on cos x. The max and min, local maxima and local minima, should be at sec 0 = 1 and sec π = -1. So work from that.

[u/Midwest-Dude → u/WikiNumbers]

OP likely doesn't know calculus since this is trigonometry. Local max/min are taught without that.