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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.



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


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.


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


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

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.



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



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.



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



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 ͡°).

Apr 20, 2019
Apr 20, 2019

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