Graphing Tangent

Trigonometry - Graphing Tangent - How it Works - Video

Letters

Now this model represents sine and cosine, but the letters are the same for the graph of tangent. With tangent, there is not an amplitude since the range is all real numbers.

Also, a helps us to determine if we have a vertical shrink or vertical stretch. If |a| > 1, then we have a vertical stretch. If |a| < 1, then we have a vertical shrink.

And, b helps us to determine if we have horizontal shrink or horizontal stretch. If |b| > 1, then we have a horizontal shrink. If |b| < 1, then we have a horizontal stretch.

Theorems

Example 1

Example 1:

Here we labeled out parts.

a = 1/3 b = 1 c = π/2 d = 0

Remember a helps us to determine if we have a vertical stretch or a vertical shrink. We have a vertical stretch since |a| > 1.

Remember b helps us find the period => π / |b| = π / |1| = π and we have a horizontal shrink since |b| > 1 .

Remember c helps us find the phase shift => -c/b = - (π /2) / 1 = -π /2, so we have phase shift to the left.

Remember d helps us find the vertical shift, since d = 0, we don't have a vertical shift.

Here we use inequality, -π/2 ≤ bx + c ≤ π/2 , to find one cycle.

a = 1/3 b = 1 c = π/2 d = 0

-π/2 < bx + c < π/2

-π/2 < 1x + π/2 < π/2

-π/2 - π/2 < x < π/2 - π/2

-π < x < 0


Substituted for b and c.

Subtracted π/2 to each section.

Added each section.

Now we have the starting point of one cycle -π and the end point 0.

Since we are dealing with tangent, this numbers represent the vertical asymptotes. Another hint is that the inequality does not have less than or equal to just less than.

Here we have graph y = tan(x) and y = (1/3) * tan(x). Knowing the parent graph of tan(x) will help graph any changes that will happen.

The black line represents y = tan(x) and the blue line represents y = (1/3) * tan(x).

We have a vertical shrink because our a value, 1/3, is less than 1. If we take a minute that makes sense since we are multiply the result of tangent by a fraction less than 1 so it takes longer to get to a larger number.

So above the x-axis the blue line is below the black line and above the x-axis the blue line is above the black line.

Here we have added another equation y = (1/3) * tan(x + π/2), which is represented by the green line.

Since c is not zero, we have moved the equation to the right or left. As there is a plus sign we have moved the equation left, because we now need a negative number to move back to the original position.

As our b = 1, our period has not changed so we can move the vertical asymptotes and the blue line by the phase shift -π/2 or to the left to make the green line.

Here we deleted the other equations that led up to y = (1/3) * tan(x + π/2).

Example 2

Example 2:

Here we labeled out parts.

a = 2 b = 4/5 c = π/4 d = -1

Remember a helps us to determine if we have a vertical stretch or a vertical shrink. We have a vertical stretch since |a| > 1.

Remember b helps us find the period => π / |b| = π / |4/5| =5π/4 and we have a horizontal shrink since |b| > 1 .

Remember c helps us find the phase shift => -c/b = - (π /4) / (4/5) = -5π /16, so we have phase shift to the left.

Remember d helps us find the vertical shift, since d = -1, we have a vertical shift down 1.

Here we use inequality, -π/2 ≤ bx + c ≤ π/2 , to find one cycle.

a = 2 b = 4/5 c = π/4 d = -1

-π/2 < bx + c < π/2

-π/2 < (4/5)x + π/4 < π/2

-π/2 - π/4 < (4/5)x < π/2 - π/4

-3π/4 < x < π/4

-15π/16 < x < 5π/16


Substituted for b and c.

Subtracted π/4 to each section.

Added each section.

Multiplied the multiplicative inverse to each section

Now we have the starting point of one cycle -15π/16 and the end point 5π/16.

Since we are dealing with tangent, this numbers represent the vertical asymptotes. Another hint is that the inequality does not have less than or equal to just less than.

Here we have graph y = tan(x), y = 2 * tan(x), and y = 2 * tan(4x/5). Knowing the parent graph of tan(x) will help graph any changes that will happen.

The black line represents y = tan(x); the blue line represents y = 2 * tan(x); the orangish line represents y = 2 * tan(4x5).

Let's take a look at the black graph and the blue graph.

We have a vertical stretch with our blue graph because our a value, 2, is greater than 1. If we take a minute that makes sense since we are multiply the result of tangent by a fraction greater than 1 so it takes a shorter time to get to a larger number.

So above the x-axis the blue line is above the black line and below the x-axis the blue line is above the black line.

Let's take a look at the blue graph and the orangish graph.

If we take a look at the middle 3 graphs, the organish graph stretches horizontal since |b| > 1. If we take a minute that makes sense since we are multiply the input value by a fraction less than 1 so it takes a longer time to get to the same number without the fraction.

The next step is to move the vertical asymptotes to match the graph.

Here we have incorporated the phase shift for the first time. Since inside the parenthesis, y = 2 * tan(4x/5+ π/4) - 1, we have a plus sign, we shift the equation horizontally to the left as now a negative number is needed to get back to zero just like the first number in our interval, -15π/16 ≤ x ≤ 5π/16.

Any point on the greenish graph directly to the left of the organish graph is 5π/16 units from each other.

Now we have to worry about the letter d. Since d = -1 our next step is to move the graph down 1 since it is positive, which represent the purple graph.

Here we have deleted the other graphs and are just left the purple one, y = 2 * tan(4x/5+ π/4) - 1, our answer.

Live Worksheet

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