I missed a few lessons and I need help:

Can someone give me an example with steps to help me understand how to find Factored/Simplified Equation, Removable Discontinuity, Vertical and Horizontal Asymptote, domain/range, and graph for an equation in the format of g(x)=something/something

I have a few examples from my homework, but if we're not allowed to have y'all do it for me, just give me an example so I won't be plagarizing.

Here they are:

g(x)=1/x^2-4

h(x)=x+3/x^2+5x+4

h(x)=x^2-9/(x+3)

f(x)=2x+1/x

y=x^2+8x+15/x+5

TheDerivativ Apr 20, 2019

#4**+3 **

__Factored/Simplified Expressions__

You take an expression, find all its factors, write it out as a product of factors, you get the factored form.

You expand all the product, then you get the simplified form.

**Example:**

\(x^2+2x-15\qquad \text{This is}\color{blue}{\text{ simplified}}\color{black}\text{.}\\ (x+5)(x-3)\qquad \text{This is}\color{magenta}{\text{ factored}}\color{black}\text{.}\\\)

__Removable discontinuity__

This usually only happens when the given function is rational.

1. Find the G.C.D. of the numerator and the denominator.

2. Set the G.C.D. to 0

3. The roots of the equation in step 2 is where the removable discontinuity occurs.

**Example:**

Find the removable discontinuity(discontinuities) of the function \(f(x) = \dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}\).

__Solution:__

\(f(x)\\ =\dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}\\ =\dfrac{\dfrac{1}{4}(x^2-4x+4)}{(x-2)(x+2)}\\ =\dfrac{\dfrac{1}{4}(x-2)^2}{(x-2)(x+2)}\\ =\dfrac{(x-2)^2}{4(x-2)(x+2)}\)

GCD of numerator and denominator is (x-2). <- Set this to 0.

x - 2 = 0

x = 2

The removable discontinuity of this function occurs at x = 2.

**Vertical asymptotes and horizontal asymptotes**

Vertical asymptotes occurs at (unremovable) discontinuities.

Horizontal asymptotes is just a horizontal line which the function approaches when x reaches positive or negative infinity.

How to find...

Vertical asymptotes? Find all the discontinuities, exclude the removable ones and the "broken" points(like this one below, at x = 1.)

Horizontal asymptotes? That's a calculus thing. If you want to do it the algebra way, you can substitute larger and larger numbers into the function and see if the function reaches a value eventually, and then repeat with negative numbers.

**Example:**

Find all the horizontal and vertical asymptotes of the function y = 1/x.

__Solution:__

Let's find the horizontal asymptotes first. If you substitute large numbers, you will find that it approaches 0 eventually. Same for negative numbers. So the **only** horizontal asymptote, in this case, is y = 0.

The function has a discontinuity at x = 0, and it is not removable. So vertical asymptote is at x = 0.

__Domain and range__

Domain is the set of possible inputs of the function.

Range is the set of possible outputs of the function.

**Example:**

Find the domain and range of the function \(f(x) = \dfrac{1}{\sqrt{x-5}}\).

__Solution:__

This requires some kind of sense of mathematics. Firstly, the denominator can't be 0, or it will give infinity. So we have \(\sqrt{x-5} \neq 0 \implies x - 5 \neq 0\).

Also, \(\sqrt{ }\) can't take negative numbers as input. So \(x - 5 \geq 0\). But actually as the denominator can't be 0, x - 5 can't be 0. So \(x - 5 > 0\implies x > 5\).

So the domain of the function \(\text{D}\{ f(x)\} \) is \((5,\infty)\).

Now let's consider the output of the function. \(\sqrt{ }\) only gives out nonnegative numbers(0 and positive numbers if you don't know what are nonnegative numbers), So \(\sqrt{x-5} \ge 0 \implies f(x) = \dfrac{1}{\sqrt{x-5}} \le \infty\), which doesn't make any sense. But \(f(x) = \dfrac{1}{\text{a nonnegative number}}\), considering that, we get the range of f(x) as the set of all positive real numbers.

\(\text{R}\{f(x)\} \) is \((0,\infty)\).

P.S.: How do I know these without knowing all the terminologies? Google it ( ͡° ͜ʖ ͡°).

MaxWong Apr 20, 2019

#2**0 **

By the way, I'm not looking for yall to answer these homework questions for me I just want a step by step method on how I would be able to work them out myself. Khan Academy is useless to me and can go die in a hole. Someone pls help and explain :/

TheDerivativ Apr 20, 2019

#3**-2 **

If Khan Academy is useless to you then you will never learn anything on here.

**Quit fucking around!**

Guest Apr 20, 2019

#4**+3 **

Best Answer

__Factored/Simplified Expressions__

You take an expression, find all its factors, write it out as a product of factors, you get the factored form.

You expand all the product, then you get the simplified form.

**Example:**

\(x^2+2x-15\qquad \text{This is}\color{blue}{\text{ simplified}}\color{black}\text{.}\\ (x+5)(x-3)\qquad \text{This is}\color{magenta}{\text{ factored}}\color{black}\text{.}\\\)

__Removable discontinuity__

This usually only happens when the given function is rational.

1. Find the G.C.D. of the numerator and the denominator.

2. Set the G.C.D. to 0

3. The roots of the equation in step 2 is where the removable discontinuity occurs.

**Example:**

Find the removable discontinuity(discontinuities) of the function \(f(x) = \dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}\).

__Solution:__

\(f(x)\\ =\dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}\\ =\dfrac{\dfrac{1}{4}(x^2-4x+4)}{(x-2)(x+2)}\\ =\dfrac{\dfrac{1}{4}(x-2)^2}{(x-2)(x+2)}\\ =\dfrac{(x-2)^2}{4(x-2)(x+2)}\)

GCD of numerator and denominator is (x-2). <- Set this to 0.

x - 2 = 0

x = 2

The removable discontinuity of this function occurs at x = 2.

**Vertical asymptotes and horizontal asymptotes**

Vertical asymptotes occurs at (unremovable) discontinuities.

Horizontal asymptotes is just a horizontal line which the function approaches when x reaches positive or negative infinity.

How to find...

Vertical asymptotes? Find all the discontinuities, exclude the removable ones and the "broken" points(like this one below, at x = 1.)

Horizontal asymptotes? That's a calculus thing. If you want to do it the algebra way, you can substitute larger and larger numbers into the function and see if the function reaches a value eventually, and then repeat with negative numbers.

**Example:**

Find all the horizontal and vertical asymptotes of the function y = 1/x.

__Solution:__

Let's find the horizontal asymptotes first. If you substitute large numbers, you will find that it approaches 0 eventually. Same for negative numbers. So the **only** horizontal asymptote, in this case, is y = 0.

The function has a discontinuity at x = 0, and it is not removable. So vertical asymptote is at x = 0.

__Domain and range__

Domain is the set of possible inputs of the function.

Range is the set of possible outputs of the function.

**Example:**

Find the domain and range of the function \(f(x) = \dfrac{1}{\sqrt{x-5}}\).

__Solution:__

This requires some kind of sense of mathematics. Firstly, the denominator can't be 0, or it will give infinity. So we have \(\sqrt{x-5} \neq 0 \implies x - 5 \neq 0\).

Also, \(\sqrt{ }\) can't take negative numbers as input. So \(x - 5 \geq 0\). But actually as the denominator can't be 0, x - 5 can't be 0. So \(x - 5 > 0\implies x > 5\).

So the domain of the function \(\text{D}\{ f(x)\} \) is \((5,\infty)\).

Now let's consider the output of the function. \(\sqrt{ }\) only gives out nonnegative numbers(0 and positive numbers if you don't know what are nonnegative numbers), So \(\sqrt{x-5} \ge 0 \implies f(x) = \dfrac{1}{\sqrt{x-5}} \le \infty\), which doesn't make any sense. But \(f(x) = \dfrac{1}{\text{a nonnegative number}}\), considering that, we get the range of f(x) as the set of all positive real numbers.

\(\text{R}\{f(x)\} \) is \((0,\infty)\).

P.S.: How do I know these without knowing all the terminologies? Google it ( ͡° ͜ʖ ͡°).

MaxWong Apr 20, 2019