Transforming general graphs

We learned how to transform the graphs of power functions $x^n$ and the trigonometric function $\sin(x)$ :

\begin{array}{lll} f(x) &=& A(x-v)^n+b\\ g(x) &=& A\sin(u(x-v))+b \end{array}

where the parameters $A, u, v$ and $b$ tells you by how much the graph of $x^n$ or $\sin(x)$ are stretched in $y$ -direction (by $A$ ), then stretched in $x$ -direction (by $1/u$ ), then shifted left or right (by $v$ ), and finally shifted up or down (by $b$ ).

Note 1

As a side remark, note that we did not use the parameter $u$ ( $x$ -direction) for the power function. We could do this, and then it would look as follows:

f(x)=A(u(x-v))^n+b

The reason why this is not necessary is that we can always write

f(x)=A(u(x-v))^n+b=Au^n(x-v)^n+b=A^\prime (x-v)^n

that is, we can replace stretching in $x$ -direction by $1/u$ by the transformation stretch in $y$ -direction by $u^n$ . Think about it: draw the graph of $x^2$ and stretch in in $x$ -direction by the factor of $4$ - it gets wider. Let's call this graph $g$ . But it is easy to find a value $A$ to stretch the graph of $x^2$ in $y$ -direction to obtain the same graph: $A=\frac{1}{16}$ . Indeed, for $f(x)=\frac{1}{16}x^2$ we get the same graph $g$ , as a table of values quickly shows.

We can transform any function in this way. We will discuss this in a moment, but let's first revisit the notation of a function, as this will be essential for what follows.

Note 2

Consider a function $f$ , say

f(x)=x^2-4\cdot x+1

Recall that this notation defines a machine. The $x$ wrapped in the brackets after the $f$ stands for the input of the machine $f$ . The algebraic formula $x^2-4\cdot x+1$ determines what to do with the input to produce the output of the machine. So perhaps a more intuitive notation would be to use a container instead of the letter $x$ , like

f(\square)=\square^2-4\cdot \square+1

and whatever is in the container in $f(\square)$ appears also in the containers in the formula. This container notation might resolve a lot of confusing issues surrounding the notation of functions, but alas, we tend not to use this notation. So instead of

f(\boxed{1})=\boxed{1}^2-4\cdot \boxed{1}+1 = -2

we simply write

f(1)=1^2-4\cdot 1+1 = -2

Importantly, if we write $f(x)= the placeholder (or variable)$ x $indicates where the input appears in the formula$ x^2-4x+1 $. The meaning of this notation is that _every$ x $in the formula is **replaced** with the thing in the brackets$ f(..)$ to dertermine the output_. For example,

\begin{array}{lll} f(1) &=& 1^2-4\cdot 1+1 = -2\\ f(\text{house}) &=& \text{house}^2-4\cdot \text{house}+1= ??\\ f(x-1) &=& (x-1)^2-4\cdot (x-1)+1 = x^2-6x+6\\ f(\sin(x))&=&\sin(x)^2-4\cdot \sin(x)+1\\ f(\square) &=& \square^2-4\cdot \square +1 \end{array}

Notice that

in the third example we have to wrap the input $x-1$ into brackets in te formula, so that the input is squared: we cannot write $f(x-1)=x-1^2-4\cdot x-1+1$ . The way we treat $+$ and $\cdot$ , this would not square the input $x-1$ , only $1$ , and it would not multiply the input $x-1$ by $4$ , only $x$ .
some inputs lead to new functions, e.g. $f(x-1)=x^2-6x+6$ , so we could write $g(x)=f(x-1)=x^2-6x+6$
for some inputs, like house, the formula does not make sense in terms of adding and multiplying, but this is okey.

Let's assume we have a starting function $f$ , whose graph is shown below. Let's apply each type of transformation to the graph of $f$ and see what the function equation of the resulting graph looks like.

Stretching in $y$ -direction by factor $A$ .