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f ( x ) = A tan ( B x C ) + D

The graph of a transformed tangent function is different from the basic tangent function tan x in several ways:

Features of the graph of y = A Tan( Bx C )+ D

  • The stretching factor is | A | .
  • The period is π | B | .
  • The domain is x C B + π | B | k , where k is an integer.
  • The range is ( −∞ , ) .
  • The vertical asymptotes occur at x = C B + π 2 | B | k , where k is an odd integer.
  • There is no amplitude.
  • y = A tan ( B x ) is and odd function because it is the qoutient of odd and even functions(sin and cosine perspectively).

Given the function y = A tan ( B x C ) + D , sketch the graph of one period.

  1. Express the function given in the form y = A tan ( B x C ) + D .
  2. Identify the stretching/compressing factor , | A | .
  3. Identify B and determine the period, P = π | B | .
  4. Identify C and determine the phase shift, C B .
  5. Draw the graph of y = A tan ( B x ) shifted to the right by C B and up by D .
  6. Sketch the vertical asymptotes, which occur at   x = C B + π 2 | B | k , where   k   is an odd integer.
  7. Plot any three reference points and draw the graph through these points.

Graphing one period of a shifted tangent function

Graph one period of the function y = −2 tan ( π x + π ) −1.

  • Step 1. The function is already written in the form y = A tan ( B x C ) + D .
  • Step 2. A = −2 , so the stretching factor is | A | = 2.
  • Step 3. B = π , so the period is P = π | B | = π π = 1.
  • Step 4. C = π , so the phase shift is C B = π π = −1.
  • Step 5-7. The asymptotes are at x = 3 2 and x = 1 2 and the three recommended reference points are ( −1.25 , 1 ) , ( −1, −1 ) , and ( −0.75, −3 ) . The graph is shown in [link] .
    A graph of one period of a shifted tangent function, with vertical asymptotes at x=-1.5 and x=-0.5.
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How would the graph in [link] look different if we made A = 2 instead of −2 ?

It would be reflected across the line y = 1 , becoming an increasing function.

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Given the graph of a tangent function, identify horizontal and vertical stretches.

  1. Find the period P from the spacing between successive vertical asymptotes or x -intercepts.
  2. Write f ( x ) = A tan ( π P x ) .
  3. Determine a convenient point ( x , f ( x ) ) on the given graph and use it to determine A .

Identifying the graph of a stretched tangent

Find a formula for the function graphed in [link] .

A graph of two periods of a modified tangent function, with asymptotes at x=-4 and x=4.
A stretched tangent function

The graph has the shape of a tangent function.

  • Step 1. One cycle extends from –4 to 4, so the period is P = 8. Since P = π | B | , we have B = π P = π 8 .
  • Step 2. The equation must have the form f ( x ) = A tan ( π 8 x ) .
  • Step 3. To find the vertical stretch A , we can use the point ( 2 , 2 ) .
    2 = A tan ( π 8 2 ) = A tan ( π 4 )

Because tan ( π 4 ) = 1 , A = 2.

This function would have a formula f ( x ) = 2 tan ( π 8 x ) .

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Find a formula for the function in [link] .

A graph of four periods of a modified tangent function, Vertical asymptotes at -3pi/4, -pi/4, pi/4, and 3pi/4.

g ( x ) = 4 tan ( 2 x )

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Analyzing the graphs of y = sec x And y = csc x

The secant    was defined by the reciprocal identity sec x = 1 cos x . Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at π 2 , 3 π 2 , etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

We can graph y = sec x by observing the graph of the cosine function because these two functions are reciprocals of one another. See [link] . The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
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bill
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-24m+3+3mÁ^2
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
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Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
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given that (7x-5):(2+4x)=8:7find the value of x
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3x-12y=18
Kelvin
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
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d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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