Graphing Tangent and Cotangent Lesson

In the last lesson, we discussed how to sketch the graphs for sine and cosine. We learned that sine and cosine are periodic functions with a period of 2$π$.
The tangent and cotangent functions are also periodic functions, however, the period in each case is $π$. $$\tan \hspace{.1em}x=\tan(x + πn)$$ $$\cot \hspace{.1em}x=\cot(x + πn)$$ Where n is any integer and x is a member of the domain for the given function. $$\tan\left(\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{4}+ \pi\right)$$ $$\tan\left(\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{4}+ \frac{4\pi}{4}\right)$$ $$\tan\left(\frac{\pi}{4}\right)=\tan\left(\frac{5\pi}{4}\right)$$ $$1=1$$ $$\cot\left(\frac{2\pi}{3}\right)=\cot\left(\frac{2\pi}{3}+ \pi\right)$$ $$\cot\left(\frac{2\pi}{3}\right)=\cot\left(\frac{2\pi}{3}+ \frac{3\pi}{3}\right)$$ $$\cot\left(\frac{2\pi}{3}\right)=\cot\left(\frac{5\pi}{3}\right)$$ $$-\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{3}$$

Tangent Function

$$\tan \hspace{.1em}x=\frac{\sin \hspace{.1em}x}{\cos \hspace{.1em}x}$$ The tangent values are 0 when sine values are 0. This happens at an x-value of 0, $π$, 2$π$, 3$π$,... ($π$n, where n is any integer). Since division by zero is undefined, the tangent function is undefined where cosine values are 0. This occurs at $π$/2, 3$π$/2, 5$π$/2,... (odd multiples of $π$/2).
A vertical asymptote is a vertical line that the graph approaches but does not touch. As the x-values get closer and closer to the line, the function values will increase or decrease without bound (approach positive or negative infinity). We can approximate the value of $π$/2 to be about 1.570796. Be careful with rounding, you may end up with something larger than $π$/2. Most books will just use 1.5707 (using truncation to avoid the problem). A quick look at a table of values for y = tan x shows what happens as x gets close to $π$/2.

y = tan x
x	y
$0$	$0$
$\large\frac{\pi}{6}$	$≈0.58$
$\large\frac{\pi}{4}$	$1.00$
$\large\frac{\pi}{3}$	$≈1.73$
$1.5$	$≈14.10$
$1.53$	$≈24.50$
$1.57$	$≈1{,}255.77$
$1.5707$	$≈10{,}381.33$
$1.570796$	$≈3{,}060{,}023.31$
$\large\frac{\pi}{2}$	undefined

As x approaches $π$/2 from the left, the value of tan x increases without bound. In other words, we could say that tan x approaches infinity as x approaches $π$/2 from the left. Additionally, as x approaches -$π$/2 from the right, the value of tan x decreases without bound. Again, we could also rephrase this and say tan x approaches negative infinity as x approaches -$π$/2 from the right. Our graph of y = tan x will approach the vertical lines x = -$π$/2 and x = $π$/2. These vertical lines are vertical asymptotes.
Let's take the information we know so far and create a first sketch of our tangent graph over one period. We can use a table for the basic points found over the interval from $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$:

y = tan x
x	y
$\large-\frac{\pi}{2}$	undefined
$\large-\frac{\pi}{3}$	$\large-\sqrt{3}$
$\large-\frac{\pi}{4}$	$\large-1$
$\large-\frac{\pi}{6}$	$\large-\frac{\sqrt{3}}{3}$
$\large0$	$\large0$
$\large\frac{\pi}{6}$	$\large\frac{\sqrt{3}}{3}$
$\large\frac{\pi}{4}$	$\large1$
$\large\frac{\pi}{3}$	$\large\sqrt{3}$
$\large\frac{\pi}{2}$	undefined

Graphing the Tangent Function over One Period

Desmos link for more detail From the graph above, we can see that as x increases from $-π$/2 to $π$/2 (not including either), tangent values range from negative infinity (-∞) to positive infinity (∞) and increase throughout the interval. The same values are repeated as x increases from $π$/2 to 3$π$/2 (not including either), and so on. It is only necessary to graph y = tan x over an interval of length $π$. The remainder of the graph consists of repetitions of that graph at intervals of $π$.
Desmos link for more detail

The Tangent Curve and its Characteristics

• The domain contains all real numbers except odd multiples of $π$/2
• f(x) = tan x is discontinuous at odd multiples of $π$/2
  • The graph contains vertical asymptotes at those values
• The x-intercepts are of the form x = $π$n, where n is any integer
  • The x-intercept occurs at 1/2 of the way between consecutive vertical asymptotes
    • Using y = tan x over the interval: $\left(-\frac{π}{2}, \frac{π}{2}\right)$:
    • $(0,0)$
  • The point on the graph 1/4 of the way between consecutive vertical asymptotes has a y-coordinate of -1
    • Using y = tan x over the interval: $\left(-\frac{π}{2}, \frac{π}{2}\right)$:
    • $\left(-\frac{π}{4},-1\right)$
  • The point on the graph 3/4 of the way between consecutive vertical asymptotes has a y-coordinate of 1
    • Using y = tan x over the interval: $\left(-\frac{π}{2}, \frac{π}{2}\right)$:
    • $\left(\frac{π}{4},1\right)$
• The period is $π$
• There are no minimum or maximum values
  • The range is: $(-\infty, \infty)$
  • The graph has no amplitude
• The graph is symmetric with respect to the origin
  • tan(-x) = -tan x, for all x in the domain
  • f(x) = tan x is an odd function

Graphing Variations of the Tangent Function

$$y=a\tan(bx - c), b > 0$$

1. Find two consecutive vertical asymptotes by finding an interval that contains one period:
   • $$-\frac{π}{2}< bx - c < \frac{π}{2}$$
2. Find a pair of consecutive vertical asymptotes:
   • $$bx - c=-\frac{π}{2}$$
   • $$bx - c=\frac{π}{2}$$
3. Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes:
   • Find the period as $π$/b
   • Find the quarter period as $π$/4b
   • Use the quarter period to find the x-values for the 1/4, 1/2, and 3/4 points
   • Use graphing transformations or evaluate the function to find the y-values
4. Join the points with a smooth curve, approaching each vertical asymptote.
5. (Optional) Repeat the steps for additional cycles to the left or right as needed.

Example #1: Find the period, phase shift, vertical shift, and two consecutive vertical asymptotes. Sketch the graph. $$f(x)=2\tan\left(2x + \frac{2π}{3}\right) + 1$$ Before we begin our problem, most students find it easier to work in factored form. This is optional but makes finding the phase shift much easier. $$f(x)=2\tan\left[2\left(x + \frac{π}{3}\right)\right] + 1$$ $$\text{Phase Shift}: \text{left}\frac{π}{3}$$ $$\text{Period}: \frac{π}{2}$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}1$$ Steps 1 & 2) Find two consecutive vertical asymptotes:
We do this by finding an interval that contains one period. $$-\frac{π}{2}< 2\left(x + \frac{π}{3}\right) < \frac{π}{2}$$ Multiply both sides by 1/2: $$-\frac{π}{4}< x + \frac{π}{3}< \frac{π}{4}$$ Add -$π/3$ to each part: $$-\frac{π}{4}-\frac{π}{3}< x < \frac{π}{4}-\frac{π}{3}$$ $$-\frac{3π}{12}-\frac{4π}{12}< x < \frac{3π}{12}-\frac{4π}{12}$$ $$-\frac{7π}{12}< x < -\frac{π}{12}$$ An interval containing one period: $$\left(-\frac{7π}{12}, -\frac{π}{12}\right)$$ Two consecutive vertical asymptotes occur at: $$x=-\frac{7π}{12}$$$$\text{and}$$$$x=-\frac{π}{12}$$ Step 3) Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes.
We previously found the period using the formula $π$/b: $$\text{Period}: \frac{π}{2}$$ Alternatively, we can subtract our x-values from the consecutive vertical asymptotes to obtain the period: $$-\frac{π}{12}- \left(-\frac{7π}{12}\right)$$ $$-\frac{π}{12}+ \frac{7π}{12}=\frac{6π}{12}=\frac{π}{2}$$ Q.P. is 1/4 of the period: $$\frac{π}{2}\cdot \frac{1}{4}=\frac{π}{8}$$ Starting with the x-value from the leftmost consecutive vertical asymptote, add the quarter period to find the x-value for the 1/4 point, 1/2 point, and 3/4 point.
1/4 Point (x-value): $$-\frac{7π}{12}+ \frac{π}{8}$$ $$-\frac{14π}{24}+ \frac{3π}{24}=-\frac{11π}{24}$$ 1/2 Point (x-value): $$-\frac{11π}{24}+ \frac{π}{8}$$ $$-\frac{11π}{24}+ \frac{3π}{24}=-\frac{8π}{24}=- \frac{π}{3}$$ 3/4 Point (x-value): $$-\frac{π}{3}+ \frac{π}{8}$$ $$-\frac{8π}{24}+ \frac{3π}{24}=-\frac{5π}{24}$$ We have our x-values for the 1/4, 1/2, and 3/4 points. We can get the y-values by plugging into the function. This is very slow and tedious. The alternative approach is to use our knowledge of graphing transformations. $$f(x)=2\tan\left(2x + \frac{2π}{3}\right) + 1$$ When compared to our basic tangent function, the graph here will be vertically stretched by a factor of 2 and shifted up by 1 unit. We must make sure to do the transformations in the correct order. Multiply by 2 first, and then add 1.
Start with the y-value for the 1/4 point of the basic tangent function, which is -1: $$-1 \cdot 2 + 1=-1$$ Next, the y-value for the 1/2 point of the basic tangent function is 0 (x-int): $$0 \cdot 2 + 1=1$$ Finally, the y-value for the 3/4 point of the basic tangent function is 1: $$1 \cdot 2 + 1=3$$ Our 1/4, 1/2, and 3/4 points: $$\left(-\frac{11π}{24}, -1\right)$$ $$\left(-\frac{π}{3}, 1\right)$$ $$\left(-\frac{5π}{24}, 3\right)$$ Steps 4 & 5) Let's sketch the graph. Desmos link for more detail Example #2: Find the period, phase shift, vertical shift, and two consecutive vertical asymptotes. Sketch the graph. $$f(x)=\tan\left(\frac{x}{2}- \frac{5π}{4}\right) + 1$$ Before we begin our problem, most students find it easier to work in factored form. This is optional but makes finding the phase shift much easier. $$f(x)=\tan\left[\frac{1}{2}\left(x - \frac{5π}{2}\right)\right] + 1$$ $$\text{Phase Shift}: \text{right}\frac{5π}{2}$$ $$\text{Period}: 2π$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}1$$ Steps 1 & 2) Find two consecutive vertical asymptotes:
We do this by finding an interval that contains one period. $$-\frac{π}{2}< \frac{1}{2}\left(x - \frac{5π}{2}\right) < \frac{π}{2}$$ Multiply both sides by 2: $$-π < x - \frac{5π}{2}< π$$ Add $5π/2$ to each part: $$-π + \frac{5π}{2}< x < π+ \frac{5π}{2}$$ $$-\frac{2π}{2}+ \frac{5π}{2}< x < \frac{2π}{2}+ \frac{5π}{2}$$ $$\frac{3π}{2}< x < \frac{7π}{2}$$ An interval containing one period: $$\left(\frac{3π}{2}, \frac{7π}{2}\right)$$ Two consecutive vertical asymptotes occur at: $$x=\frac{3π}{2}$$$$\text{and}$$$$x=\frac{7π}{2}$$ Step 3) Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes.
We previously found the period using the formula $π$/b: $$\text{Period}: 2π$$ Alternatively, we can subtract our x-values from the consecutive vertical asymptotes to obtain the period: $$\frac{7π}{2}- \frac{3π}{2}=\frac{4π}{2}=2π$$ Q.P. is 1/4 of the period: $$\frac{2π}{4}=\frac{π}{2}$$ Starting with the x-value from the leftmost consecutive vertical asymptote, add the quarter period to find the x-value for the 1/4 point, 1/2 point, and 3/4 point.
1/4 Point (x-value): $$\frac{3π}{2}+ \frac{π}{2}=\frac{4π}{2}=2π$$ 1/2 Point (x-value): $$2π + \frac{π}{2}=\frac{4π}{2}+ \frac{π}{2}=\frac{5π}{2}$$ 3/4 Point (x-value): $$\frac{5π}{2}+ \frac{π}{2}=\frac{6π}{2}=3π$$ We have our x-values for the 1/4, 1/2, and 3/4 points. We can get the y-values by plugging into the function. This is very slow and tedious. The alternative approach is to use our knowledge of graphing transformations. $$f(x)=\tan\left(\frac{x}{2}- \frac{5π}{4}\right) + 1$$ When compared to our basic tangent function, the graph will be shifted up by 1 unit.
Start with the y-value for the 1/4 point of the basic tangent function, which is -1: $$-1 + 1=0$$ Next, the y-value for the 1/2 point of the basic tangent function is 0 (x-int): $$0 + 1=1$$ Finally, the y-value for the 3/4 point of the basic tangent function is 1: $$1 + 1=2$$ Our 1/4, 1/2, and 3/4 points: $$\left(2π, 0\right)$$ $$\left(\frac{5π}{2}, 1\right)$$ $$\left(3π, 2\right)$$ Steps 4 & 5) Let's sketch the graph. Desmos link for more detail

Graphing the Cotangent Function

$$\cot\hspace{.1em} x=\frac{\cos \hspace{.1em} x}{\sin \hspace{.1em} x}$$ We can use an almost identical process to graph the cotangent function. We just need to understand a few differences between the functions. The period for the cotangent function is also $π$. Since cosine is now in the numerator and sine is now in the denominator, the cotangent values are 0 when the cosine values are 0, and are undefined when sine values are 0. This means we will have x-intercepts at odd multiplies of $π$/2, and vertical asymptotes at $π$n, where n is any integer. As x increases from 0 to $π$ (not including either), the cotangent values range from positive infinity (∞) to negative infinity (-∞) and decrease throughout the interval. Those same values are repeated as x increases from $π$ to 2$π$ (not including either), and so on.
Let's create a first sketch of our cotangent graph over one period. We can use a table for the basic points found over the interval from $\left(0, π\right)$:

y = cot x
x	y
$\large0$	undefined
$\large\frac{\pi}{6}$	$\large\sqrt{3}$
$\large\frac{\pi}{4}$	$\large1$
$\large\frac{\pi}{3}$	$\large\frac{\sqrt{3}}{3}$
$\large\frac{\pi}{2}$	$\large0$
$\large\frac{2π}{3}$	$\large-\frac{\sqrt{3}}{3}$
$\large\frac{3\pi}{4}$	$\large-1$
$\large\frac{5\pi}{6}$	$\large-\sqrt{3}$
$\largeπ$	undefined

Graphing the Cotangent Function over One Period

Desmos link for more detail Just like with the tangent function, the remainder of the graph consists of repetitions of that graph at intervals of $π$.
Desmos link for more detail

The Cotangent Curve and its Characteristics

• The domain contains all real numbers except $π$n, where n is any integer
• f(x) = cot x is discontinuous at $π$n, where n is any integer
  • The graph contains vertical asymptotes at those values
• The x-intercepts are of the form x = $\frac{π}{2}+ πn$
  • The x-intercept occurs at 1/2 of the way between consecutive vertical asymptotes
    • Using y = cot x over the interval: $\left(0, π\right)$:
    • $\left(\frac{π}{2},0\right)$
  • The point on the graph 1/4 of the way between consecutive vertical asymptotes has a y-coordinate of 1
    • Using y = cot x over the interval: $\left(0, π\right)$:
    • $\left(\frac{π}{4},1\right)$
  • The point on the graph 3/4 of the way between consecutive vertical asymptotes has a y-coordinate of -1
    • Using y = cot x over the interval: $\left(0, π\right)$:
    • $\left(\frac{3π}{4},-1\right)$
• The period is $π$
• There are no minimum or maximum values
  • The range is: $(-\infty, \infty)$
  • The graph has no amplitude
• The graph is symmetric with respect to the origin
  • cot(-x) = -cot x, for all x in the domain
  • f(x) = cot x is an odd function

Graphing Variations of the Cotangent Function

$$y=a\cot(bx - c), b > 0$$

1. Find two consecutive vertical asymptotes by finding an interval that contains one period:
   • $$0 < bx - c < π$$
2. Find a pair of consecutive vertical asymptotes:
   • $$bx - c=0$$
   • $$bx - c=π$$
3. Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes:
   • Find the period as $π$/b
   • Find the quarter period as $π$/4b
   • Use the quarter period to find the x-values for the 1/4, 1/2, and 3/4 points
   • Use graphing transformations or evaluate the function to find the y-values
4. Join the points with a smooth curve, approaching each vertical asymptote.
5. (Optional) Repeat the steps for additional cycles to the left or right as needed.

Example #3: Find the period, phase shift, vertical shift, and two consecutive vertical asymptotes. Sketch the graph. $$f(x)=\cot\left(\frac{x}{2}+ \frac{π}{3}\right)$$ Before we begin our problem, most students find it easier to work in factored form. This is optional but makes finding the phase shift much easier. $$f(x)=\cot\left[\frac{1}{2}\left(x + \frac{2π}{3}\right)\right]$$ $$\text{Phase Shift}: \text{left}\frac{2π}{3}$$ $$\text{Period}: 2π$$ $$\text{Vertical Shift}: \text{None}$$ Steps 1 & 2) Find two consecutive vertical asymptotes:
We do this by finding an interval that contains one period. $$0 < \frac{1}{2}\left(x + \frac{2π}{3}\right) < π$$ Multiply both sides by 2: $$0 < x + \frac{2π}{3}< 2π$$ Add -$2π/3$ to each part: $$0 - \frac{2π}{3}< x < 2π - \frac{2π}{3}$$ $$- \frac{2π}{3}< x < \frac{6π}{3}- \frac{2π}{3}$$ $$- \frac{2π}{3}< x < \frac{4π}{3}$$ An interval containing one period: $$\left(-\frac{2π}{3}, \frac{4π}{3}\right)$$ Two consecutive vertical asymptotes occur at: $$x=-\frac{2π}{3}$$$$\text{and}$$$$x=\frac{4π}{3}$$ Step 3) Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes.
We previously found the period using the formula $π$/b: $$\text{Period}: 2π$$ Alternatively, we can subtract our x-values from the consecutive vertical asymptotes to obtain the period: $$\frac{4π}{3}- \left(-\frac{2π}{3}\right)=\frac{6π}{3}=2π$$ Q.P. is 1/4 of the period: $$\frac{2π}{4}=\frac{π}{2}$$ Starting with the x-value from the leftmost consecutive vertical asymptote, add the quarter period to find the x-value for the 1/4 point, 1/2 point, and 3/4 point.
1/4 Point (x-value): $$-\frac{2π}{3}+ \frac{π}{2}$$ $$-\frac{4π}{6}+ \frac{3π}{6}=-\frac{π}{6}$$ 1/2 Point (x-value): $$-\frac{π}{6}+ \frac{π}{2}$$ $$-\frac{π}{6}+ \frac{3π}{6}=\frac{2π}{6}=\frac{π}{3}$$ 3/4 Point (x-value): $$\frac{π}{3}+ \frac{π}{2}$$ $$\frac{2π}{6}+ \frac{3π}{6}=\frac{5π}{6}$$ We have our x-values for the 1/4, 1/2, and 3/4 points. We can get the y-values by plugging into the function. This is very slow and tedious. The alternative approach is to use our knowledge of graphing transformations. $$f(x)=\cot\left(\frac{x}{2}+ \frac{π}{3}\right)$$ When compared to our basic cotangent function, the graph has no vertical transformations.
Use the y-values for the 1/4, 1/2, and 3/4 points of the basic cotangent function. This gives us y-values of 1, 0, and -1, respectively.
Our 1/4, 1/2, and 3/4 points: $$\left(-\frac{π}{6}, 1\right)$$ $$\left(\frac{π}{3}, 0\right)$$ $$\left(\frac{5π}{6}, -1\right)$$ Steps 4 & 5) Let's sketch the graph. Desmos link for more detail Example #4: Find the period, phase shift, vertical shift, and two consecutive vertical asymptotes. Sketch the graph. $$f(x)=\frac{1}{2}\cot\left(2x + \frac{2π}{3}\right) + 2$$ Before we begin our problem, most students find it easier to work in factored form. This is optional but makes finding the phase shift much easier. $$f(x)=\frac{1}{2}\cot\left[2\left(x + \frac{π}{3}\right)\right] + 2$$ $$\text{Phase Shift}: \text{left}\frac{π}{3}$$ $$\text{Period}: \frac{π}{2}$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}2$$ Steps 1 & 2) Find two consecutive vertical asymptotes:
We do this by finding an interval that contains one period. $$0 < 2\left(x + \frac{π}{3}\right) < π$$ Multiply both sides by 1/2: $$0 < x + \frac{π}{3}< \frac{π}{2}$$ Add -$π/3$ to each part: $$0 - \frac{π}{3}< x < \frac{π}{2}- \frac{π}{3}$$ $$-\frac{π}{3}< x < \frac{3π}{6}- \frac{2π}{6}$$ $$-\frac{π}{3}< x < \frac{π}{6}$$ An interval containing one period: $$\left(-\frac{π}{3}, \frac{π}{6}\right)$$ Two consecutive vertical asymptotes occur at: $$x=-\frac{π}{3}$$$$\text{and}$$$$x=\frac{π}{6}$$ Step 3) Find the points on the graph at 1/4, 1/2, and 3/4 of the way between the consecutive vertical asymptotes.
We previously found the period using the formula $π$/b: $$\text{Period}: \frac{π}{2}$$ Alternatively, we can subtract our x-values from the consecutive vertical asymptotes to obtain the period: $$\frac{π}{6}- \left(-\frac{π}{3}\right)$$ $$\frac{π}{6}+ \left(\frac{2π}{6}\right)=\frac{3π}{6}=\frac{π}{2}$$ Q.P. is 1/4 of the period: $$\frac{π}{2}\cdot \frac{1}{4}=\frac{π}{8}$$ Starting with the x-value from the leftmost consecutive vertical asymptote, add the quarter period to find the x-value for the 1/4 point, 1/2 point, and 3/4 point.
1/4 Point (x-value): $$-\frac{π}{3}+ \frac{π}{8}$$ $$-\frac{8π}{24}+ \frac{3π}{24}=-\frac{5π}{24}$$ 1/2 Point (x-value): $$-\frac{5π}{24}+ \frac{π}{8}$$ $$-\frac{5π}{24}+ \frac{3π}{24}=-\frac{2π}{24}=-\frac{π}{12}$$ 3/4 Point (x-value): $$-\frac{π}{12}+ \frac{π}{8}$$ $$-\frac{2π}{24}+ \frac{3π}{24}=\frac{π}{24}$$ We have our x-values for the 1/4, 1/2, and 3/4 points. We can get the y-values by plugging into the function. This is very slow and tedious. The alternative approach is to use our knowledge of graphing transformations. $$f(x)=\frac{1}{2}\cot\left(2x + \frac{2π}{3}\right) + 2$$ When compared to our basic cotangent function, the graph here will be vertically shrunk by a factor of 1/2 and shifted up by 2 units. We must make sure to do the transformations in the correct order. Multiply by 1/2 first, and then add 2.
Start with the y-value for the 1/4 point of the basic cotangent function, which is 1: $$1 \cdot \frac{1}{2}+ 2=\frac{1}{2}+ 2$$ $$\frac{1}{2}+ \frac{4}{2}=\frac{5}{2}$$ Next, the y-value for the 1/2 point of the basic cotangent function is 0 (x-int): $$0 \cdot \frac{1}{2}+ 2=2$$ Finally, the y-value for the 3/4 point of the basic cotangent function is -1: $$-1 \cdot \frac{1}{2}+ 2=-\frac{1}{2}+ 2$$ $$-\frac{1}{2}+ \frac{4}{2}=\frac{3}{2}$$ Our 1/4, 1/2, and 3/4 points: $$\left(-\frac{5π}{24}, \frac{5}{2}\right)$$ $$\left(-\frac{π}{12}, 2\right)$$ $$\left(\frac{π}{24}, \frac{3}{2}\right)$$ Steps 4 & 5) Let's sketch the graph. Desmos link for more detail

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