Linear functions are order-1 polynomials and can be informally thought about as the graph of lines if their input is a real number. A related concept are linear mappings which are equivalent transformations on matrices. More about this family of functions on [libretexts](https://k12.libretexts.org/Bookshelves/Mathematics/Analysis/01%3A_Analyzing_Functions/1.04%3A_Function_Families/1.4.01%3A_Linear_and_Absolute_Value_Function_Families) ![[Linear function example.png]] File available on [geogebra](https://www.geogebra.org/calculator/vbpfgqhm) ## Slope The slope is $m=\frac{\Delta y}{\Delta x} = \frac{\text{rise}}{\text{run}}$*Note:* 1. All lines $y_n = mx_n + c_n$ (ie with the same $m$) are parallel 2. All lines with $m_2 =-\frac{1}{m_1}$ are perpendicular to lines with slope $m_1$ 3. if the sign of $m$ is positive, then $f(x)$ will _increase_ as $x$ increases so the graph will go _up_ as $x$ goes to the right. $f(x)$ is increasing everywhere ie over the interval $(-\infty,\infty)$ 4. If $m$ is negative then $f(x)$ will _decrease_ as $x$ increases so the graph will go _down_ as $x$ goes to the right. $f(x)$ is decreasing everywhere ie over the interval $(-\infty,\infty)$ 5. A common case is where you have two points $(x_0, y_0)$ and $(x_1, y_1)$ and need to calculate their slope. In this case it is simply $ m = \frac{y_0 - y_1}{x_0 - x_1}$ Although you can hopefully see that this is just another way to state the equation above, this form is helpful to just know as it is how point-slope form is derived. ## Slope intercept form A helpful form of any linear function is _slope intercept_ form, given as $f(x) = mx +c$ where $m$ is the slope and $\left(0,c\right)$ is the y-intercept. The $x$-intercept can be found by setting $y=0$ in the equation of the line and solving for $x$. $ \begin{align*} 0 &= mx+c \\ mx &= -c \\ x &= -\frac{c}{m} \end{align*} $ So the $x$-intercept of an arbitrary linear function $f(x) = mx+c$ is at $\left(-\frac{c}{m},0\right).$ ## Point slope form The _point slope_ form is an alternative form for the equation of a line and is given as $f(x) = m(x-x_0) + y_0$ where $(x_0, y_0)$ is some point that has been given. This can be helpful for problems where: - we are given $m$ directly - we are told the line is parallel to some other line. Since parallel lines have the same slope this is equivalent to being told $m$. - the line is perpendicular to some other line. Lines perpendicular to some line $y=mx + c_1$ have the form $y=-\frac{1}{m}x+c_2$ so this is also equivalent to being told the slope. The point we are given could be any arbitrary point, so the intersection of two lines or the $y$ or $x$ intercept would all qualify for this treatment. The slope-intercept form could be seen as actually a special case of the point slope form with the point being the y-intercept $(0,c)$. ### Derivation of point-slope form The most important slope equation comes if you have 2 points $(x,y)$ and $(x_0, y_0)$.[^1] Then the respective deltas are just the difference of the coordinates, so $ \begin{align*} m &= \frac{y-y_0}{x-x_0} & \text{...which means} \\ y-y_0 &= m(x-x_0) & \text{...and thence}\\ y &=m(x-x_0) + y_0, & \text{which is point slope form.} \end{align*} $ [^1]: I'm using the slightly weird naming because it makes the derivations below more obvious because they work without variable renaming. ## Obtaining the graph of a linear function by translation The graph of a linear function can be obtained by translation from the graph of $f(x) = x$ via the following set of steps that never need to change 1. Rewrite into slope intercept form 2. if $m$ is negative, mirror $f(x)$ about the $x$ axis 3. $|m|$ is the factor of vertical scaling, so take the graph obtained from step 2 and scale it vertically by $|m|$ [^1] 4. $c$ is the vertical shift. If $c$ is positive, shift _up_ by $c$, if $c$ is negative, shift _down_ by $|c|$.[^2] [^1]: It can be seen that for linear functions, scaling vertically by $m$ is equivalent to scaling horizontally by $\frac{1}{m}$. So stretching $x$ is the same as squashing $y$ and vice-versa [^2]: The point-slope form should make clear that for linear functions, shifting $m$ units vertically is the same as shifting 1 unit horizontally. ## Inverse of a linear function Given some arbitrary linear $f(x) = mx + c \quad f:\mathbf{R} \to \mathbf{R}$ with $m,c \in \mathbf{R}, m\neq 0$, we can find $f^{-1}(x)$ (its inverse) by substituting $x$ for $f(x)$ and $g$ for $x$ and solving for $g$ as follows: $ \begin{align*} x &= mg + c \\ mg &= x -c \\ g &= \frac{1}{m}x - \frac{c}{m}. \end{align*} $ We can see that if $f$ has type $\mathbf{R} \to \mathbf{R}$ and $m$ and $c$ are real numbers (with $m\neq0$)that the same would be true about $g$ and also that $g$ should be defined everywhere $f$ is defined. Let us also check whether the function $g(x) = \frac{1}{m}x - \frac{c}{m}$ we have found is indeed the inverse of $f(x) = mx+c$. If this is the case, then $f\circ g(x) = x$ and $g\circ f(x) = x$ So let us evaluate those by substitution: $ \begin{align*} f\circ g(x) &= m\left( \frac{1}{m}x - \frac{c}{m}\right) + c \\ &= \frac{m}{m} x -\frac{mc}{m} + c\\ &= x - c + c\\ &= x\\[12pt] g\circ f(x) &= \frac{1}{m}\left( mx + c\right) - \frac{c}{m} \\ &= \frac{m}{m} x +\frac{c}{m} - \frac{c}{m} \\ &= x \end{align*} $ Therefore the inverse of any real-valued linear function in slope intercept form $f(x) = mx+c$ with $m,c\in \mathbf{R}$ and $m\neq0$ is $f^{-1}(x)=\frac{1}{m}x - \frac{c}{m}.$ The only remaining step to finding the inverse of any arbitrary linear function would be if the domain of $f(x)$ is restricted for some reason, apply the corresponding restriction to the domain of $g(x)$ by evaluating $f$ at the boundary points of its domain (this will give you the image set of $f$) and using that interval to bound the domain of $g$. The image set of $g$ will then be the restriction applied to the domain of $f$. *Note:* 1. The graphs of $f(x)$ and $f^{-1}(x)$ are symmetrical about $y=x$. 2. If $f(x)$ is linear then $f^{-1}(x)$ is linear and vice versa. 3. If $m=0$ the function is not invertible. If you think about it, if it existed the inverse of something like $f(x)=c$ would have a domain of $c$ and map it to every possible $x$, which would be "interesting". ### Obtaining the graph of $f^{-1}(x)$ from $y=x$ by translation As with any other linear function, you could just carry out the steps above. 1. Mirror about the $x$ axis if $m$ is negative 2. Scale vertically by $\frac{1}{|m|}$ 3. Shift vertically by $-\frac{c}{m}$ ## Special cases 1. $x=c$ is not a function (because it not a “one to one” or “many to one” mapping, so fails the “vertical line test”) so has no inverse. It seems related to linear functions given that it does define a line, it's just the line has infinite slope. For the same set of reasons its limit is not defined anywhere, and it is not continuous anywhere. 2. $f(x)=c$ is a function with $m=0$. It is constant and continuous everywhere and has limit $c$ everywhere (incl at $\pm\infty$ - see [[Limits and continuity of linear functions (incl proofs)#|here]] for proofs). It is not one to one so does not have an inverse however. ## Parametric form of a linear function A straight line can be expressed in _parametric form_ as a set of point vectors $\{P + tA\} \qquad t\in \mathbf{R}. \tag{1}$ In this form $t$ is a scalar parameter. $A$ is a vector of coefficients giving the slope with respect to $t$ in each dimension and $p$ is an offset equivalent to the $c$ parameter in the normal slope intercept form. Both $A$ and $P$ are vectors, so what this is really saying is you obtain the line by starting with some vector $P$ and adding $A$ times the scalar factor $t$. On a point of notation, these are written in uppercase because they are point vectors, rather than writing them $\mathbf{a}$ and $\mathbf{p}$ or similar. A conceptually-simpler version of this which possibly appears more frequently is a set of parametric equations such as $ \begin{align*} x &= at + b\\ y &= ct + d \end{align*} $ With $a\ne 0, c\ne 0$. As a general matter, any parametric system of the form $\begin{align*} x &= a_1t + p_1\\ y &= a_2t + p_2\\ \vdots \\ x_n &= a_nt + p_n \end{align*} $ with $a_1 \ne 0, \dots a_n \ne 0$ describes a linear function parameterised by $t$. Written like this it, is hopefully easy to see that no matter how many dimensions you have, you can scoop up all the coefficients of $t$ into a vector and call it $A$ and take all the constants and call those $P$, and you have the form in (1). ## Piecewise linear functions TODO ## Absolute value of a linear function TODO ## Limits and proofs of continuity Are in their own page because it was getting kinda long and unwieldy. [[Limits and continuity of linear functions (incl proofs)]]