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education/math/MATH1050/Dividing Polynomials.md

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+# Long division
+$$ \frac{6x^2-x-2}{2x+1} $$
+Given the above problem, you'd first start by figuring out what you'd multiple the largest exponent in the denominator ($2x$) by, to equal the largest number in the numerator ($6x^2$). In this case, that number is $3x$. You'd then multiple the entire denominator by that number, giving you $6x^2 + 3x$. $3x$ would then be part of your solution, and you'd subtract $6x^2+3x$ from $6x^2-x-2$ to give you $-4x-2$. The process is then repeated with $-4x-2$, figuring out what you'd multiply to cancel it out, then adding that to your solution. This is repeated until you can't anymore. This is your remainder, and can be written as $\frac{r}{2x+1}$, where $r$ is your remainder, and $2x + 1$ is the denominator in the original equation.
+
+# Synthetic Division
+This is a slightly more efficient method of devision that's valid when the denominator is in the form of $x \pm n$, or $x$ plus or minus a number.
+
+https://en.wikipedia.org/wiki/Synthetic_division

education/math/MATH1050/Domain.md

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+# Examples
+
+Given the below problem, the two equations can't simplified further. So to find the domain, you need to look for the domain where they're both valid, eg $[-2, 5]$.
+$$ \sqrt{x+2} + \sqrt{5-x} $$
+The below example has a domain of $[-2, 5)$ because $x$ cannot equal 0 for the denominator
+$$ \frac{\sqrt{x+2}}{\sqrt{5-x}} $$
+Assuming $f(x) = \frac{2}{x-3}$, and $g(x) = \frac{5}{x+1}$, $(f\circ g)(x)$, you can find the domain by finding the domain for each function, then fully expanding it and seeing if any more unreachable numbers are included 

education/math/MATH1050/Exponents/Exponents.md

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+The standard form of an exponential function looks something like this:
+$$ a(b)^{cx-h}+k $$
+- $a$: Vertical stretch/compression
+- $c$ Horizontal stretch/compression
+- $h$: Horizontal translation left or right
+- $k$: Vertical translation up or down
+
+Without stretch:
+$$ a^{x-b}+c $$
+
+A negative exponent is the equivalent of `1/x`, EG
+$$ x^{-2} = \frac{1}{x^2} $$
+An equation in the base form $2^x$ will have an asymptote of $y = 0$. 
+
+Putting a negative in front of something like $2^x$ flips it over the *x axis* ($-2^x$).
+Putting a negative in front of the exponent ($2^{-x}$) flips it over the y axis. 
+
+## Solving Exponents
+To solve an equation that's got variables in the exponents, work on setting the bases on each side equal, then you can treat the exponents like standalone. Refer to the notes on [[Logarithms]] for ways to do that. 

education/math/MATH1050/Exponents/Logarithms.md

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+ https://www.youtube.com/watch?v=sULa9Lc4pck
+
+$$log_a(b) $$
+Pronounced log *base* a, this function is used to figure out what exponent you need to raise $a$ to to get $b$.
+
+$log_ab = c$ can be rewritten as $a^c = b$.
+
+
+$$ 5^{log_5^{(x+2)}}=x+2 $$
+By default, $log$ refers to $log_{10}$. $ln$ is shorthand for $log_e$. 
+
+$$ \sqrt{x} = x^{1/2} $$
+To get the reciprocal of a value, change the sign of the exponent.
+$$ x^{-1} = \frac{1}{x} $$
+## Domain
+There are 3 places you need to worry about domain. The third is specific to logarithms.
+- You can't divide by 0
+- You can't take the square root of a negative without complex numbers
+- You cannot take the $log$ of a zero, or a negative number.
+	- There's no way to raise a number to an exponent and have it equal zero, or be a negative number.
+	- This can be used to help solve inequalities, because you know *an equation that's wrapped in a logarithm must be $> 0$*.
+
+### Finding the domain of added logarithms
+$$ log(x+2) + log(2x-3) $$
+With the above example, you can find the domain of each function separately, then find the overlap of valid numbers.
+
+
+# Expanding logarithms
+$$ log_b(x*y) = log_b x + log_b y $$
+$$log_b(\frac{x}{y}) = log_b x - log_b y $$
+
+Example Problem:
+$$ log_5 z = 3 $$
+$$ log_5 y = 2 $$$$log_5(yz) = log_5 y + log_5 z $$$$ 2 + 3 = 5 $$
+Exponents can be moved to the front of a logarithm
+$$ log_3 x^5 = 5*log_3 x $$
+Roots are just the inverse, so:
+$$ log_3 \sqrt x = \frac{1}{2}*log_3x $$
+## Change of base
+$$ log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b} $$
+The above are all equivalent because the ratios are the same
+
+
+### The compound interest formula
+$$ A= Pe^{rt} $$
+
+| Value | Description |
+| ---- | ---- |
+| $A$ | Ending amount |
+| $P$ or $A_0$ | Starting amount |
+| $r$ or $k$ | Rate (a %) |
+| $t$ | The amount of times interest is compounded |

education/math/MATH1050/Exponents/Quadratics.md

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+# Notes
+
+To convert an equation in the form of $y = (x^2-3x)$ into a square equivalent, you:
+- Take the second value, $3$, then half it, giving you ($\frac{3}{2}$). 
+- Then to rebalance the equation, you're going to square that value (giving you $\frac{9}{4}$), then add it to the other side, multiplying it by $a$ if necessary, where $a$ is what the parentheses are multiplied by in the form $a(x - h)^2$. This will give you an equation that looks something like: $\frac{9}{4} + y = (x -\frac{3}{2})^2$.
+- Finally, you can rebalance the equation by subtracting $\frac{9}{4}$ from both sides, giving you $y = (x - \frac{3}{2})^2 - \frac{9}{4}$. This equation should be equal to the original.
+
+$$ y = -5x^2 -20x + 13 $$
+Given the above equation, you can factor out a -5, resulting in the equation $-5(x^2+4x) + 13)$. Half of 4 is 2, and because the inside is multiplied by -5, $-5 *4 = -20$, so you add -20 to the other side to equalize the equation, resulting in an equation in the form of $-20 + y = -5(x+2)^2+ 13$. This simplifies down to $y = -5(x+2)^2 + 33$. 
+# Forms
+**Standard form (vertex form)**
+$$ y = a(x - h)^2 + k $$
+To convert to standard form given a vertex of a quadratic equation and a point that falls along that line, plug values in for everything and solve for $a$. 
+
+**Quadratic form**
+$$ y = a^2 + bx + c $$
+# End Behavior of functions
+If the largest exponent of a function is **even**, both sides of a function will point the same way.
+Given there's no negative coefficient, as $x$ goes towards infinity, $f(x)$ will go towards infinity. As $f(x)$ goes towards $-\infty$, $f(x)$ will still go to infinity. A negative coefficient will flip this.
+$$ x \rightarrow \infty, \space f(x) \rightarrow \infty $$
+
+$$ x \rightarrow -\infty, \space f(x) \rightarrow -\infty $$
+If the largest exponent of a function is **odd**, each side of the function will point towards a different direction. Given there's no negative coefficient, the left side of the graph will point down, and the right side will point up. A negative coefficient will flip this.
+
+The *least degree* of a polynomial is the number of turning points + 1. 

education/math/MATH1050/Inverse Functions.md

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+For a function to have an inverse, it needs to have one $x$ for every $y$, and vice versa. You can use the horizontal line test to verify that the inverse of a function is valid. If you can draw a horizontal line and it crosses through two points at the same time at any height, the inverse is not a valid function. To get the inverse, you can switch the x and y of a function, and it will mirror the graph over the line $y = x$. 
+
+# Examples
+Given the below function:
+$$ y = \frac{1}{2}x + 3 $$
+You can find the inverse by switching the $x$ and $y$ values and solving for $y$:
+$$ x = \frac{1}{2}y + 3 $$
+The range of the inverse is the same as the domain of the original. 
+You can verify by taking $f \circ g$, and simplifying. 
+

education/math/MATH1050/Matrices.md

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+$$ 
+\begin{bmatrix}
+1 & 2 & 3\\
+4 & 5 & 6
+\end{bmatrix}
+$$
+To write a system of equations as a matrix, you use the coefficients of each term. The solutions for each equation might be written with a bar separating:
+
+$$ 
+\begin{bmatrix}
+1 & 2 & 3|4\\
+4 & 5 & 6|5
+\end{bmatrix}
+$$
+## Multiplying matrices
+To multiply two matrices, the *number of columns* for the first matrix must match the *number of rows* in the second.
+
+For example: the following matrices could be multiplied together because the number of columns in the first matrix (3) matches the number of rows in the second matrix (3).
+
+$$ 
+\begin{bmatrix}
+1 & 2 & 3\\
+4 & 5 & 6
+\end{bmatrix} *
+\begin{bmatrix}
+1 & 2\\
+3 & 4\\
+5 & 6
+\end{bmatrix}
+$$
+
+
+When multiplying two matrices together, you're going to go row by column, just like the order. You'll multiply the first entry in the row of the first matrix by the first entry in the column of the second matrix, then proceed along, doing this for the second entry, and so on, making a list of numbers. With the above matrix, and the first row/column, it looks like this:
+$${1*1,2*3,3*5}$$
+With the second row/column, it looks like this:
+$${4*2,5*4,6*6}$$
+The list is added together.
+
+The resulting value is inserted into the cell where the row/column intercept.
+
+%% To get answers, go through each entry in the second matrix, entry by entry, and multiply by every item in the matching row of the first one, then add all of those together. %%
+
+https://www.youtube.com/watch?v=XkY2DOUCWMU
+
+## Adding matrices
+To add two matrices together, they must be the same dimensionally (same number of rows, same number of columns).

education/math/MATH1050/Partial Fractions.md

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+Partial fraction decomposition is when you break a polynomial fraction down into smaller fractions that add together. 
+## Degree where the numerator is less
+This is the "main" method of solving, and the next two headings both focus on getting to this point, at which point you solve using the below steps.
+1. Factor the bottom. 
+2. Create two fractions, $\frac{a}{p1}$, and $\frac{b}{p2}$, where p1 and p2 are the polynomials you just factored out, and a/b are arbitrary variables
+3. Multiply a by p2, and b by p1., giving you: $$\frac{a*p2}{p1} + \frac{b*p1}{p2}$$
+4. Now you can distribute $a$ and $b$, giving you $ax + c$ and $bx + d$. Group and factor. Your equation equals the original equation, so the numerator of the first equals $ax + bx + c + d$.
+
+### Example
+$$ \frac{2x+1}{(x+1)(x+2)} $$
+1. Given the above fraction, the denominator is already factored, so we can move onto the next step, where two fractions are made with $a$ and $b$ in the numerator, and each of the denominator components in the denominator:
+$$ \frac{a}{x+1} + \frac{b}{x+2} $$
+2. Next, find a common denominator so that you can add the two fractions together. In this case, it's $(x+1)(x+2)$
+3. Then to make the denominators equal, you're going to multiply the numerator and the denominator by the component that the denominator is missing. In this example, the denominator for $\frac{a}{x+1}$ is missing $x+2$, so you're going to multiply by $\frac{x+2}{x+2}$, and vice versa for $\frac{b}{x+2}$:
+$$ \frac{a(x+2)}{(x+1)(x+2)} + \frac{b(x+1)}{(x+2)(x+1)} $$
+4. You can now add the two equations together and distribute $a$ and $b$, giving you:
+$$ \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
+	This equals the first equation, so:
+$$ \frac{2x+1}{(x+1)(x+2)} = \frac{ax+2a + bx+b}{(x+1)(x+2)} $$
+5. Notice that the denominator on both sides is equal, meaning you can cancel them out, giving you:
+$$ 2x + 1 = ax + 2a + bx + b $$
+6. Next, group your $x$ values on one side, and your constants on the other side. You'll notice that you can factor $ax + bx$, giving you $x(a+b)$.
+$$ 2x+1 = x(a + b) + (2a + b) $$
+7. With the above equation, each side is in the same form. it's $x$ multiplied by a constant ($2$ on the left, and $(a+b)$ on the right, and with a constant of $1$ on the left and $2a + b$) on the right, letting you find the two equations below:
+$$ 2 = a + b $$
+$$ 1 = 2a + b $$
+8. The above equations can be solved as a system of equations, giving you:
+$$ a=-1,\space b=3 $$
+9. Your answer would be:
+   $$ \frac{-1}{x+1} + \frac{3}{x+2} $$
+
+## Degree of the numerator is equal
+1. First perform polynomial division.
+2. Then find a partial fraction with the remainder
+
+## Degree where the numerator is greater
+1. First perform polynomial division to reach a point where the degree of the numerator is less than the degree of the denominator, then follow those steps for the remainder. Your answer is in the form of $a + r$, where $a$ was your answer for division, and $r$ is your reformatted remainder.

education/math/MATH1050/Polynomial Fractions.md

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+
+| Value | Instructions | Example |
+| ---- | ---- | ---- |
+| x intercept | Solve the *top of the fraction* for x | $\frac{x-1}{x+2}$ -> $x-1 = 0$ -> $x_{int} = 1$ |
+| y intercept | divide the constant term on top by the constant term on bottom | $\frac{3x+1}{2x+2}$-> $\frac{3}{2}$ |
+| vertical asymptote(s) | Set the *bottom of the fraction* to 0 and solve (find the roots) | $\frac{x-1}{x-2}$ -> $x-2 = 0$ -> $VA = 2$ |
+| Horizontal asymptote | - if the degree of the leading coefficient on the top is less than the degree on the bottom, $y = 0$.<br>- If the degree on the top equals the degree on the bottom, y = `Leading Coefficient of Top / Leading Coefficient of Bottom`.<br>- If the degree on the top is greater than the degree on the bottom, divide to find the slant/oblique asymptote.<br> | - Degree on top is smaller than degree on bottom<br>$\frac{x-1}{x^2+2}$ -> $y=0$<br>- Degree on top is the same as degree on bottom  |
+
+## Point of discontinuity
+A point of discontinuity is created when you cancel terms out of the top and the bottom, the cancelled term creates a hole in the graph. For example, if you cancelled out $x-2$, a hole would be created on the graph at $x = 2$.
+
+To solve for the y coordinate of a point of discontinuity, take the equation after it's simplified, and plug the x coordinate of the PoD into the equation. 
+
+| Term | Definition |
+| ---- | ---- |
+| Degree | The power that a variable is raised to. EG, $x^5$ would have a degree of 5 |
+| Leading Term | The element in the polynomial with the highest degree. EG, in the polynomial $3x^4 + 2x^3 + 5x^2 - 3x + 6$, $3x^4$ would be the leading term because it has the highest degree. |
+| Leading Coefficient | The coefficient of the leading term in a polynomial. For example, if the leading term was $3x^4$, the leading coefficient would be $3$. |
+| Constant Term | The number in a polynomial that is not multiplied by a variable. EG, $7$.  |

education/math/MATH1050/Rational Inequalities.md

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+## Basic Form
+$$ \frac{x+3}{x-4} < 0 $$
+
+1. Draw a number line. 
+2. Look at the bottom of the fraction $x - 4$, solve for x($x = 4$ ). When the bottom equals zero, put an empty circle on the line to mark an empty point, because you cannot divide by 0.
+3. Look at the top, solve for zero, and put another point on the line. If it's $\le 0$, this point is filled in, otherwise it's another hole. Now check each "section" along the line by plugging in an arbitrary value to see if the result evaluates to less than zero.
+
+$$ \frac{(x+1)^2(x-1)}{(x+4)(x-3)} \le 0 $$
+In this case, solving the bottom means that you'll have empty points on $x - 4$ and $x - 3$. Solving the top means that you'll have filled in points at $x = -1$ and $x = 1$. At this point the number line will be divided into chunks. You can pick an arbitrary number in each chunk and plug it in for $x$. This will let you figure out which parts of the range are valid. The result is written in the form of $(m, M) \cup (m, M)$.
+
+If the other side isn't zero (EG, $< 3$), you'll move everything to one side. You can do this by multiplying the right side by the denominator on the bottom of the fraction 
+$$ \frac{2x-17}{x-5} \lt 3 $$
+In this example, you'd multiply the right side by $\frac{x-5}{x-5}$. Once that's done, you can subtract it over and combine. 

education/math/MATH1050/Standard forms of circles.md

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+# Notes
+## Composition of functions
+For $(f\circ g)(x)$ for two sets, you look for $x$ from $f$ and an equivalent $y$ value from $g$, and leftover coordinates are the answer. The order of $f$ and $g$ does matter. 
+# Formulae
+The general equation for a circle:
+$$ (x - h)^2 + (y - k)^2 =r^2 $$
+Distance formula:
+$$ \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$
+Midpoint formula:
+$$ (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) $$
+Adding functions:
+$$ (f + g)(x) = f(x) + g(x) $$
+Multiplying functions:
+$$ (f*g)(x)=f(g(x)) $$
+
+# Examples
+Given the function $f = \{(0, 2), (3, -1), (5, 4), (2, 1)\}$, and $g=\{(2, 0), (3, -1), (4, -2), (5, 2)\}$, and applying $(f+g(x)$, If the same $x$ value exists in both the sets $f$ and $g$, then you can solve for that value by adding $y$ values for the matching x coordinates together. 
+
+
+# Terminology
+
+| Term | Definition |
+|--|--|
+| $h$ | How far left or right something is shifted from the origin |
+| $k$| How far up or down something is shifted from the origin |
+| $r$ | The radius of a circle |

education/math/MATH1050/Systems of Equations.md

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+| Term | Definition |
+| ---- | ---- |
+| Consistent | The system of equations has at least one solution |
+| Inconsistent | Parallel lines, no solution |
+| Independent | The lines only cross at one point. |
+| Dependant | The lines are identical, and there are infinitely many solutions. Both equations represent the same line when plotted. |
+# Solving
+## Graphing
+Graph the two equations, and look for points where they intersect
+
+## Substitution
+Solve for a variable in the equation, then substitute it into another equation.
+EG, given the below systems:
+$$ y = x+3 $$
+$$ x = 3y + 2 $$
+You can substitute values, like this:
+$$ x = 3(x+3) + 2 $$
+Then once you have $x$, you can plug it back into the first equation to solve for y.
+
+## Elimination
+To solve a system of equations by elimination, you add the two equations together and see if something cancels out.
+A simple example might look like this:
+$$ 2x + 3y = 23 $$
+$$ 4x -3y = -13 $$
+You "add" the two equations together, getting:
+$$ 6x + 0y = 10 $$
+You now know that $6x = 10$.
+If you don't have two values that evenly cancel out, like $3$ and $-4$, you can find the least common multiple and multiply the entire equation so that those two are equal. In this case, you'd multiply one equation by 4, and one equation by 3. If the signs don't cancel out, you can multiply one of the equations by -1.
+
+If given systems of equations with 3 variables, you pick two equations, and eliminate one variable from them. Then you pick one of the first two equations, and the last equation, and eliminate that same variable from them. This will give you two different equations that only have two variables. You can then use elimination again, and solve for one variable. This allows you to solve the rest of the equation.