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Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.
$g(x)=3x-2$
Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch ; if the constant is greater than 1, we get a horizontal compression of the function.
Given a function $\text{\hspace{0.17em}}y=f(x),\text{\hspace{0.17em}}$ the form $\text{\hspace{0.17em}}y=f(bx)\text{\hspace{0.17em}}$ results in a horizontal stretch or compression. Consider the function $\text{\hspace{0.17em}}y={x}^{2}.\text{\hspace{0.17em}}$ Observe [link] . The graph of $\text{\hspace{0.17em}}y={\left(0.5x\right)}^{2}\text{\hspace{0.17em}}$ is a horizontal stretch of the graph of the function $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ by a factor of 2. The graph of $\text{\hspace{0.17em}}y={\left(2x\right)}^{2}\text{\hspace{0.17em}}$ is a horizontal compression of the graph of the function $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ by a factor of 2.
Given a function $\text{\hspace{0.17em}}f(x),\text{\hspace{0.17em}}$ a new function $\text{\hspace{0.17em}}g(x)=f(bx),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is a constant, is a horizontal stretch or horizontal compression of the function $\text{\hspace{0.17em}}f(x).$
Given a description of a function, sketch a horizontal compression or stretch.
Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, $\text{\hspace{0.17em}}R,\text{\hspace{0.17em}}$ will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.
Symbolically, we could write
See [link] for a graphical comparison of the original population and the compressed population.
A function $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ is given as [link] . Create a table for the function $\text{\hspace{0.17em}}g(x)=f\left(\frac{1}{2}x\right).$
$x$ | 2 | 4 | 6 | 8 |
$f(x)$ | 1 | 3 | 7 | 11 |
The formula $\text{\hspace{0.17em}}g(x)=f\left(\frac{1}{2}x\right)\text{\hspace{0.17em}}$ tells us that the output values for $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are the same as the output values for the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at an input half the size. Notice that we do not have enough information to determine $\text{\hspace{0.17em}}g(2)\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}g(2)=f\left(\frac{1}{2}\cdot 2\right)=f(1),\text{\hspace{0.17em}}$ and we do not have a value for $\text{\hspace{0.17em}}f(1)\text{\hspace{0.17em}}$ in our table. Our input values to $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ will need to be twice as large to get inputs for $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ that we can evaluate. For example, we can determine $\text{\hspace{0.17em}}g(4)\text{.}$
We do the same for the other values to produce [link] .
$x$ | 4 | 8 | 12 | 16 |
$g(x)$ | 1 | 3 | 7 | 11 |
[link] shows the graphs of both of these sets of points.
Relate the function $\text{\hspace{0.17em}}g(x)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ in [link] .
The graph of $\text{\hspace{0.17em}}g(x)\text{\hspace{0.17em}}$ looks like the graph of $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ horizontally compressed. Because $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ ends at $\text{\hspace{0.17em}}(6,4)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g(x)\text{\hspace{0.17em}}$ ends at $\text{\hspace{0.17em}}(2,4),\text{\hspace{0.17em}}$ we can see that the $\text{\hspace{0.17em}}x\text{-}$ values have been compressed by $\text{\hspace{0.17em}}\frac{1}{3},\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}6\left(\frac{1}{3}\right)=2.\text{\hspace{0.17em}}$ We might also notice that $\text{\hspace{0.17em}}g(2)=f\left(6\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g(1)=f\left(3\right).\text{\hspace{0.17em}}$ Either way, we can describe this relationship as $\text{\hspace{0.17em}}g(x)=f\left(3x\right).\text{\hspace{0.17em}}$ This is a horizontal compression by $\text{\hspace{0.17em}}\frac{1}{3}.$
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