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When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch?
When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression?
A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.
When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x -axis from a reflection with respect to the y -axis?
How can you determine whether a function is odd or even from the formula of the function?
For a function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ substitute $\text{\hspace{0.17em}}(-x)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}(x)\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}f(x).\text{\hspace{0.17em}}$ Simplify. If the resulting function is the same as the original function, $\text{\hspace{0.17em}}f(-x)=f(x),\text{\hspace{0.17em}}$ then the function is even. If the resulting function is the opposite of the original function, $\text{\hspace{0.17em}}f(-x)=-f(x),\text{\hspace{0.17em}}$ then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.
Write a formula for the function obtained when the graph of $\text{\hspace{0.17em}}f(x)=\sqrt{x}\text{\hspace{0.17em}}$ is shifted up 1 unit and to the left 2 units.
Write a formula for the function obtained when the graph of $\text{\hspace{0.17em}}f(x)=\left|x\right|\text{\hspace{0.17em}}$ is shifted down 3 units and to the right 1 unit.
$g(x)=|x-1|-3$
Write a formula for the function obtained when the graph of $\text{\hspace{0.17em}}f(x)=\frac{1}{x}\text{\hspace{0.17em}}$ is shifted down 4 units and to the right 3 units.
Write a formula for the function obtained when the graph of $\text{\hspace{0.17em}}f(x)=\frac{1}{{x}^{2}}\text{\hspace{0.17em}}$ is shifted up 2 units and to the left 4 units.
$g(x)=\frac{1}{{(x+4)}^{2}}+2$
For the following exercises, describe how the graph of the function is a transformation of the graph of the original function $\text{\hspace{0.17em}}f.$
$y=f(x-49)$
$y=f(x+43)$
The graph of $\text{\hspace{0.17em}}f(x+43)\text{\hspace{0.17em}}$ is a horizontal shift to the left 43 units of the graph of $\text{\hspace{0.17em}}f.$
$y=f(x+3)$
$y=f(x-4)$
The graph of $\text{\hspace{0.17em}}f(x-4)\text{\hspace{0.17em}}$ is a horizontal shift to the right 4 units of the graph of $\text{\hspace{0.17em}}f.$
$y=f(x)+5$
$y=f(x)+8$
The graph of $\text{\hspace{0.17em}}f(x)+8\text{\hspace{0.17em}}$ is a vertical shift up 8 units of the graph of $\text{\hspace{0.17em}}f.$
$y=f(x)-2$
$y=f(x)-7$
The graph of $\text{\hspace{0.17em}}f(x)-7\text{\hspace{0.17em}}$ is a vertical shift down 7 units of the graph of $\text{\hspace{0.17em}}f.$
$y=f(x-2)+3$
$y=f(x+4)-1$
The graph of $f(x+4)-1$ is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of $f.$
For the following exercises, determine the interval(s) on which the function is increasing and decreasing.
$f(x)=4{(x+1)}^{2}-5$
$g(x)=5{(x+3)}^{2}-2$
decreasing on $\text{\hspace{0.17em}}(-\infty ,-3)\text{\hspace{0.17em}}$ and increasing on $\text{\hspace{0.17em}}(-3,\infty )$
$a(x)=\sqrt{-x+4}$
$k(x)=-3\sqrt{x}-1$
decreasing on $(0,\text{\hspace{0.17em}}\infty )$
For the following exercises, use the graph of $\text{\hspace{0.17em}}f(x)={2}^{x}\text{\hspace{0.17em}}$ shown in [link] to sketch a graph of each transformation of $\text{\hspace{0.17em}}f(x).$
$g(x)={2}^{x}+1$
$w(x)={2}^{x-1}$
For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions.
$h(x)=|x-1|+4$
$m(t)=3+\sqrt{t+2}$
Tabular representations for the functions $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ are given below. Write $\text{\hspace{0.17em}}g(x)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h(x)\text{\hspace{0.17em}}$ as transformations of $\text{\hspace{0.17em}}f(x).$
$x$ | −2 | −1 | 0 | 1 | 2 |
$f(x)$ | −2 | −1 | −3 | 1 | 2 |
$x$ | −1 | 0 | 1 | 2 | 3 |
$g(x)$ | −2 | −1 | −3 | 1 | 2 |
$x$ | −2 | −1 | 0 | 1 | 2 |
$h(x)$ | −1 | 0 | −2 | 2 | 3 |
$g(x)=f(x-1),\text{\hspace{0.17em}}h(x)=f(x)+1$
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