<< Chapter < Page | Chapter >> Page > |
$$af\left(bx+c\right)+d;\phantom{\rule{1em}{0ex}}a,b,c\in R$$
In this case "a", "b", "c" and "d" can be either positive or negative depending on the particular transformation. A positive "d" means that graph is shifted up. On the other hand, we can specify constants to be positive in the following representation :
$$\pm af\left(\pm bx\pm c\right)\pm d;\phantom{\rule{1em}{0ex}}a,b,c>0$$
The form of representation appears to be cumbersome, but is more explicit in its intent. It delinks sign from the magnitude of constants. In this case, the signs preceding positive constants need to be interpreted for the nature of transformation. For example, a negative sign before c denotes right horizontal shift. It is, however, clear that both representations are essentially equivalent and their use depends on personal choice or context. This difference does not matter so long we understand the process of graphing.
In order to understand this type of transformation, we need to explore how output of the function changes as input to the function changes. Let us consider an example of functions f(x) and f(x+1). The integral values of inedependent variable are same as integral values on x-axis of coordinate system. Note that independent variable is plotted along x-axis as real number line. The integral x+1 values to the function f(x+1) - such that input values are same as that of f(x) - are shown on a separate line just below x-axis. The corresponding values are linked with arrow signs. Input to the function f(x+1) which is same as that of f(x) corresponds to x which is 1 unit smaller. It means graph of f(x+1) is same as graph of f(x), which has been shifted by 1 unit towards left. Else, we can say that the origin of plot (also x-axis) has shifted right by 1 unit.
Let us now consider an example of functions f(x) and f(x-2). Input to the function f(x-2) which is same as that of f(x) now appears 2 unit later on x-axis. It means graph of f(x-2) is same as graph of f(x), which has been shifted by 2 units towards right. Else, we can say that the origin of plot (also x-axis) has shifted left by 2 units.
The addition/subtraction transformation is depicted symbolically as :
$$y=f\left(x\right)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}y=f\left(x\pm \left|a\right|\right);\phantom{\rule{1em}{0ex}}\left|a\right|>0$$
If we add a positive constant to the argument of the function, then value of y at x=x in the new function y=f(x+|a|) is same as that of y=f(x) at x=x-|a|. For this reason, the graph of f(x+|a|) is same as the graph of y=f(x) shifted left by unit “a” in x-direction. Similarly, the graph of f(x-|a|) is same as the graph of y=f(x) shifted right by unit “a” in x-direction.
1 : The plot of y=f(x+|a|); is the plot of y=f(x) shifted left by unit “|a|”.
2 : The plot of y=f(x-|a|); is the plot of y=f(x) shifted right by unit “|a|”.
We use these facts to draw graph of transformed function f(x±a) by shifting graph of f(x) by unit “a” in x-direction. Each point forming the plot is shifted parallel to x-axis (see quadratic graph showm in the of figure below). The graph in the center of left figure depicts monomial function $y={x}^{2}$ with vertex at origin. It is shifted right by “a” units (a>0) and the function representing shifted graph is $y={\left(x-a\right)}^{2}$ . Note that vertex of parabola is shifted from (0,0) to (a,0). Further, the graph is shifted left by “b” units (b>0) and the function representing shifted graph is $y={\left(x+b\right)}^{2}$ . In this case, vertex of parabola is shifted from (0,0) to (-b,0).
Notification Switch
Would you like to follow the 'Functions' conversation and receive update notifications?