In Exercises 5–12, graph two periods of the given tangent function.

y = −2 tan 1/2 x

Question

In Exercises 5–12, graph two periods of the given tangent function.

y = −2 tan 1/2 x

Accepted Answer

Hello, today we're gonna be graphing two periods of the given function. So what we are given is Y is equal to negative four tangent of one third X. Now, before we graph this function, let's take a look at the pair function of tangent of X. The parent function has a period of pie and the tangent function contains two asymptotes. The first one is located at negative pi over two and the second one is located at positive pie over two. The function has X intercept at zero comma zero. And the tangent graph has a rough snakelike structure which is going to look like this. So what we want to do in order to graph the given function is you want to first identify all the transformations that are occurring on the tangent function. The first transformation happens from the negative sign. The negative sign indicates that we are reflecting the, reflecting the graph about the X axis. So that means that we are taking the graph and flipping it about the X axis. The second transformation occurs with the absolute value of four. Since the coefficient of the absolute value of four is greater than one this tells us that there is going to be a stretch horizontally of the graph. So normally this four would tell us that there is a change in the amplitude of a trigonometric function. But tangent does not have an amplitude because the range is from negative infinity to positive infinity. The four is going to allow us to identify two points on the tangent graph that indicate the horizontal stretch. It won't give us the X values of these two points. But it will tell us that the Y values are going to be plus or minus four. And the third transformation of the graph is going to occur with the coefficient in front of the X term. See the coefficient in front of the X term is not one, it's given to us as 1/3. And this indicates that there is going to be a phase shift of the graph. What this means is that the period of the graph is going to change. So how do we identify the new period? Well, the period is given to us as pi over B and B represents the coefficient in front of the X term. So in this case here, P O B is going to be one third. And if we plug this into the period, we get pi divided by 1/3. In order to simplify this, we multiply this by the reciprocal of the denominator, the denominator is going to cancel the one and that leaves us with pi times 3/1, which is going to give us three pi. So the new period is going to be three pi units long. And now we need to identify the locations of the new asymptotes. Well, for the original tangent function, we know that the X term has to lie between the values of negative pi over two and pie over two. But in this case, here we are given one third X. So in order to get the new Asy Tos, we're going to be setting one third X between the values of negative pi over two and pie over two. And now we want to go ahead and isolate X. In order to do this, we're gonna multiply each term into inequality by three. Doing this allows us to cancel out the fraction in the middle leaving us with just X. And the new range for X is given to us as negative three pi over two and positive three pi over two. So that means that the new ascent totes for the tangent graph are going to be at negative three pi over two and three pi over two. And finally, because we have no horizontal or vertical shift, the X intercept is going to remain the same at the origin zero comma zero. So now that we know the new ascent totes, we have the X intercept and we've identified all of the transformations. Let's go ahead and plot the first period of the graph. So we're going to draw an axis. Now we know based off of the new range that the first period is going to be located at negative three pi over two. And the second ascent tote is gonna be located at positive three pi over two. The X intercept is still going to remain at the origin zero comma zero. Now let's apply the first transformation we identified earlier that the first tra transformation is going to be the reflection about the X axis. So the shape of the new tangent graph is roughly going to look like this. No, we want to identify two points that represent the horizontal stretch. Those two points are gonna be located halfway between the first Asymptote and the X intercept and halfway between the X intercept and the second. Now we know that the location of the first point does not, we don't know what the X value is going to be, but we do know that the Y value is gonna be positive for and for the second point, the same situation occurs, we don't know the X value, but we know the Y value is negative four. So how do we go about obtaining the X values for these two points? While we know that this point lies between the first Asymptote and the X intercept. So what we can do is we can add the distance between the first Asymptote which is negative three pi over two and the X intercept which is zero and divide that distance in half negative three pi over two plus zero is going to give us negative three pi over two, a negative three pi over two, divided by two is going to give us negative three pi over four. So the location of the first point is going to be negative three pi over four. Now we just need to find the location or the X value of the second point. Well, since that one lies between the X intercept and the second asom tote, we're going to add the distance between zero and positive three pi over two and divide that distance in half zero plus three pi over two is going to give us three pi over two and three pi over two divided by two is going to give us positive three pi over four. So the location of the second point is going to be positive three pi over four comma negative four. This is going to be the fully labeled version of the first period of the graph. But recall that the problem asks us to graph two periods of the graph. So what we need to do is we need to identify a third Asymptote for the function will recall that the distance for the new period is going to be three pi. So in order to get the location of the third ascent to, we're going to be adding three pi to three pi over two, adding these two values together is going to give us the value of nine pi over two. And that means that the location of the third ascent tode is going to be nine pi over two. And the graph of the second period is going to mirror the graph of the first period. So this is going to be the rough graph of the second period and the entirety of this axis is going to be the graph of the given tangent function. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 5–12, graph two periods of the given tangent function. y = −2 tan 1/2 x

Verified Solution

Key Concepts

Tangent Function

Transformations of Functions

Graphing Trigonometric Functions