# Graph Transformations

## Introduction

In this section we will learn how to graph modified versions of the functions from the library of functions. The transformations we will be focusing on are left/right shifts, up/down shifts, and vertical stretching/compression. Even though these properties hold for functions in general, we will focus on functions from the library of functions.

## Vertical shift

A vertical shift comes from taking a graph and moving every point up, or down, by some uniform amount. Algebraically, we take a point (x, f(x)) and move it to (x, f(x) + c), where c is the shift up or down. This means the function f(x) + c is a shift up by c units if c is positive, and down by c units if c is negative.

Example: Graph ${\displaystyle f(x)=x^{2}+4}$

Here we can recognize that the function from the library of functions is ${\displaystyle x^{2}}$. This tells us that the graph of f(x) will have the same shape as ${\displaystyle x^{2}}$. Since we are in the vertical shift section, we can guess that the function has been shifted vertically by some amount. Since we are adding 4 to the output of every point from ${\displaystyle x^{2}}$ the graph of ${\displaystyle x^{2}+4}$ is the graph of ${\displaystyle x^{2}}$ shifted up by 4 units.

## Horizontal Shift

Horizontal shifts are a little harder to describe algebraically. Graphically, it is exactly what you think it is. We take the graph of f(x) and move all of the points horizontally by a uniform amount. For identifying horizontal shifts start by attempting which function from the library of functions is the base function. Then observe where a point on the base function moves. I usually keep track of the y-intercept. So the approach is focused on determining what value of x is required to plug into the shifted function to get the same value that evaluating the base function at 0 is.

Example:

Graph ${\displaystyle f(x)=(x-5)^{3}}$

The first thing to notice is the base function from the library of functions is ${\displaystyle x^{3}.}$ The y-intercept is (0, 0). Now we want to find a value of x such that ${\displaystyle (x-5)^{3}=0}$. Solving this we find that we need x = 5. This means the point (0, 0) moved to (5, 0) and every point moved to the right 5 units.