Question

Hi this is a grade 11 math question and the unit is tangents. please answer either by sketching the graph and showing me how to find each property or by just using the equation find the different properties which are domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept. please show me this as an example as this is an example in my lesson but i dont know how to solve it, i keep coming up with the wrong answer for each property. thank you.

Sketch the graph y = 1/(tan(x)), and state its properties.

Answer 1

One over tanx =cotx

Answer 2

Look at the definition of the tangens function:
\[\tan(x) = \frac{\sin(x)}{\cos(x)}\] from this follows\[f(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}\]
What is the domain of this function? Well, we know cos(x) over all of R and sin(x) over all of R, the only problem we could have is if sin(x) equals 0 because then we would be dividing by zero, which we'll always want to avoid. So the domain of this function is R without the zeroes of sin(x). In other words: R without all x so that \[x ={ n \pi, n \in \mathbb{Z} } \]What will f(x) look like at these points? Well, as we approach them we will be dividing by smaller and smaller numbers which is equivalent to multiplying with bigger and bigger numbers, so our graph will climb (fall) steeper and steeper - We'll have vertical asymptotes at all those points.

The zeroes of our function f(x) are the zeroes of cos(x), all x which can be written like this: \[x = \left(\frac{1}{2} + n\right) \cdot \pi\, ;\, n \in \mathbb{Z}\]The next thing I'll discuss is periodicity. We know (Or at least I'll suppose that we konw that ;]) that the period of cos(x) and sin(x) is 2π, so we know that our function is also 2π -periodical as it is the quotient of the two. But also, we know that (look at the attachment!) sin(x + π) = -sin(x) and cos(x + π) = -cos(x). What does this implicate for our function?\[f(x + \pi) = \frac{\cos(x + \pi)}{\sin(x + \pi)} = \frac{(-1)\cos(x)}{(-1)\sin(x)} = \frac{\cos(x)}{\sin(x)} = f(x)\]
So in fact, f(x) is even π-periodical!

Now, even without having seen its graph before, I can tell how the graph of f(x) will look like. I can look at it between x = 0 and x = π and know how it looks anywhere because it's π-periodical. I know that it will climb to +∞ at x = 0, it has a vertical asymptote there. I know that i t will be 0 at x = π/2 so I'll draw a curve down to 0. Then i know that it will fall to -∞ at x = π so I'll draw a curve down to -∞. So I'll get a graph looking like the one in the attachment. I think you'll be able to do symmetry and y-interception yourself.