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When dealing with transformations in maths, we can observe mainly two types, they are Z transformation and Laplace transformation. Linear Algebra. Transformations of Functions Activity Builder by Desmos THE PARENT FUNCTION GRAPHS AND TRANSFORMATIONS! - YouTube This means that the rest of the functions that belong in this family are simply the result of the parent function being transformed. 4.2: Graphs of Exponential Functions - Mathematics LibreTexts stretched vertically by a factor of | a | if | a | > 0. Use a graphing calculator to verify your answer.. Horizontal shift 2 right. Exponential functions parent functions will each have a domain of all real numbers and a restricted range of (0, \infty). Log InorSign Up. As we have learned earlier, the linear functions parent function is the function defined by the equation, [kate]y = x[/katex] or [kate]f(x) = x[/katex]. These are the common transformations performed on a parent function: By transforming parent functions, you can now easily graph any function that belong within the same family. \(f(t)\), \(g(t)\) be the functions of time, \(t\), then, $$\mathscr{L}\left\{C_1f(t)+C_2g(t) \right\}=\mathscr{L}\left\{C_1f(t) \right\}+\mathscr{L}\left\{C_2g(t) \right\}$$, Read Also: Derivative Of sin^2x, sin^2(2x) & More, Read Also: Horizontal Asymptotes Definition, Rules & More, $$If\ \mathscr{L}\left\{f(t) \right\}=F(s)\ then\ \mathscr{L}\left\{e^{at}f(t) \right\}=F(s-a)$$, If\(\mathscr{L}\left\{f(t) \right\}=F(s),\ then\), $$\mathscr{L}\left\{f(at) \right\}=\frac{1}{a}F(\frac{s}{a})$$, $$\mathscr{L}\left\{f(\frac{t}{a}) \right\}=aF(sa)$$, $$\mathscr{L}\frac{d^n}{dt^n}\left\{f(t) \right\}=s^n\mathscr{L}\left\{f(t) \right\}-s^{n-1}f(0)-s^{n-2}f^1(0)-f^{n-1}(0)$$, $$\mathscr{L}\frac{d^1}{dt^1}\left\{f(t) \right\}=s\mathscr{L}\left\{f(t) \right\}-f(0)$$, $$\mathscr{L}\left[\int_{}^{}\int_{}^{}\int_{}^{}\int_{}^{}\int_{}^{}f(t)dt^n \right]=\frac{1}{s^n}\mathscr{L}\left\{f(t) \right\}+\frac{}{}+\frac{f^{n-1}(0)}{s^n}+\frac{f^{n-2}(0)}{s^n}++\frac{f^{1}(0)}{s}$$, $$\mathscr{L}\left\{\int_{0}^{t}f(t)dt \right\}=\frac{1}{s}\mathscr{L}\left\{f(t) \right\}+\frac{f^{1}(0)}{s}$$, If \(\mathscr{L}\left\{f(t) \right\}=F(s)\), then the Laplace Transform of \(f(t)\) after the delay of time, \(T\) is equal to the product of Laplace Transform of \(f(t)\) and \(e^{-st}\) that is, $$\mathscr{L}\left\{f(t-T)u(t-T) \right\}=e^{-st}F(s)$$.