Vertical Dilations (Stretch/Shrink)

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Question 1 of 5

1. Question
Dilate (stretch/shrink) $y = ln x$ $y=lnx$ vertically by a factor of $\frac{1}{3}$ $1/3$ .
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A dilation is to stretch or to shrink the shape of a curve.

Vertical dilations (stretch/shrink) are shown by $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $k$ is the vertical dilation factor.

If $0 < k < 1$ $0<color(red)(k)<1$ , then the graph is compressed.

If $k > 1$ $color(red)(k)>1$ , then the graph is stretched.

To dilate (stretch/shrink) $y = ln x$ $y=lnx$ vertically by a factor of $\frac{1}{3}$ $color(red)(1/3)$ , use $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $color(red)(k)$ is the vertical dilation factor. This means that $k = \frac{1}{3}$ $color(red)(k=1/3)$ .

The equation for the new dilated equation is $y = \frac{1}{3} ln x$ $y=color(red)(1/3)lnx$ .

Since $k = \frac{1}{3}$ $color(red)(k=1/3)$ then the graph is compressed vertically as it will be further away from the y-axis.

Use a table of values to find the $y$ $y$ values for both the original and dilated graphs.

$x$ $x$ $\frac{1}{2}$ $1/2$ $1$ $1$ $3$ $3$

$ln x$ $lnx$ $- 0.7$ $-0.7$ $0$ $0$ $1.1$ $1.1$

$\frac{1}{3} ln x$ $1/3lnx$ $- 0.2$ $-0.2$ $0$ $0$ $0.37$ $0.37$

Sketch the graph of $y = ln x$ $y=lnx$ and $y = \frac{1}{3} ln x$ $y=1/3lnx$ using the table of values.
Question 2 of 5

2. Question
Dilate (stretch/shrink) $y = x^{2}$ $y=x^2$ vertically by a factor of $3$ $3$ .
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A Dilation is to stretch or to shrink the shape of a curve.

Vertical dilations (stretch/shrink) are shown by $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $k$ is the vertical dilation factor.

If $0 < k < 1$ $0<color(red)(k)<1$ , then the graph is shrinked.

If $k > 1$ $color(red)(k)>1$ , then the graph is stretched.

To dilate (stretch/shrink) $y = x^{2}$ $y=x^2$ vertically by a factor of $3$ $color(red)(3)$ , use $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $color(red)(k)$ is the vertical dilation factor. This means that $k = 3$ $color(red)(k=3)$ .

$y = 3 x^{2}, k = 3$ $y=color(red)(3)x^2, color(red)(k=3)$ , therefore the graph is stretched vertically along the y-axis and it will appear narrower.

Use a table of values to find the $y$ $y$ values for both the original and dilated graphs.

$x$ $x$ $- 2$ $-2$ $- 1$ $-1$ $0$ $0$ $1$ $1$ $2$ $2$

$x^{2}$ $x^2$ $4$ $4$ $1$ $1$ $0$ $0$ $1$ $1$ $4$ $4$

$3 x^{2}$ $3x^2$ $12$ $12$ $3$ $3$ $0$ $0$ $3$ $3$ $12$ $12$

Sketch the graph of $y = x^{2}$ $y=x^2$ and $y = 3 x^{2}$ $y=3x^2$ using the table of values.
Question 3 of 5

3. Question
Dilate (stretch/shrink) $y = e^{x}$ $y=e^x$ vertically by a factor of $\frac{1}{4}$ $1/4$ .
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A dilation is to stretch or to shrink the shape of a curve.

Vertical dilations (stretch/shrink) are shown by $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $k$ is the vertical dilation factor.

If $0 < k < 1$ $0<color(red)(k)<1$ , then the graph is compressed.

If $k > 1$ $color(red)(k)>1$ , then the graph is stretched.

To dilate (stretch/shrink) $y = e^{x}$ $y=e^x$ vertically by a factor of $\frac{1}{4}$ $color(red)(1/4)$ , use $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $color(red)(k)$ is the vertical dilation factor. This means that $k = \frac{1}{4}$ $color(red)(k=1/4)$ .

The equation for the new dilated equation is $y = \frac{1}{4} e^{x}$ $y=color(red)(1/4)e^x$ .

Since $k = \frac{1}{4}$ $color(red)(k=1/4)$ the graph is compressed vertically. It is further away from the y-axis.

Use a table of values to find the $y$ $y$ values for both the original and dilated graphs.

$x$ $x$ $0$ $0$ $1$ $1$ $2$ $2$ $3$ $3$

$e^{x}$ $e^x$ $1$ $1$ $2.7$ $2.7$ $7.4$ $7.4$ $20$ $20$

$\frac{1}{4} e^{x}$ $1/4e^x$ $\frac{1}{4}$ $1/4$ $0.7$ $0.7$ $1.9$ $1.9$ $5$ $5$

Sketch the graph of $y = e^{x}$ $y=e^x$ and $y = \frac{1}{4} e^{x}$ $y=1/4e^x$ using the table of values.
Question 4 of 5

4. Question
Dilate (stretch/shrink) $y = x^{2}$ $y=x^2$ vertically by a factor of $\frac{1}{2}$ $1/2$ .
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A dilation is to stretch or to shrink the shape of a curve.

Vertical dilations (stretch/shrink) are shown by $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $k$ is the vertical dilation factor.

If $0 < k < 1$ $0<color(red)(k)<1$ , then the graph is compressed.

If $k > 1$ $color(red)(k)>1$ , then the graph is stretched.

To dilate (stretch/shrink) $y = x^{2}$ $y=x^2$ vertically by a factor of $\frac{1}{2}$ $color(red)(1/2)$ , use $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $color(red)(k)$ is the vertical dilation factor. This means that $k = \frac{1}{2}$ $color(red)(k=1/2)$ .

The equation for the new dilated equation is $y = \frac{1}{2} x^{2}$ $y=color(red)(1/2)x^2$ .

Since $k = \frac{1}{2}$ $color(red)(k=1/2)$ the graph is compressed vertically so it is further away from the y-axis.

Use a table of values to find the $y$ $y$ values for both the original and dilated graphs.

$x$ $x$ $1$ $1$ $2$ $2$ $3$ $3$ $4$ $4$

$x^{2}$ $x^2$ $1$ $1$ $4$ $4$ $9$ $9$ $16$ $16$

$\frac{1}{2} x^{2}$ $1/2x^2$ $\frac{1}{2}$ $1/2$ $2$ $2$ $\frac{9}{2}$ $9/2$ $8$ $8$

Sketch the graph of $y = x^{2}$ $y=x^2$ and $y = \frac{1}{2} x^{2}$ $y=1/2x^2$ using the table of values.
Question 5 of 5

5. Question
Dilate (stretch/shrink) $y = | x |$ $y=|x|$ vertically by a factor of $4$ $4$ .
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A dilation is to stretch or to shrink the shape of a curve.

Vertical dilations (stretch/shrink) are shown by $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $k$ is the vertical dilation factor.

If $0 < k < 1$ $0<color(red)(k)<1$ , then the graph is compressed.

If $k > 1$ $color(red)(k)>1$ , then the graph is stretched.

To dilate (stretch/shrink) $y = | x |$ $y=|x|$ vertically by a factor of $4$ $color(red)(4)$ , use $y = k f (x)$ $y=color(red)(k)f(x)$ where $k$ $color(red)(k)$ is the vertical dilation factor. This means that $k = 4$ $color(red)(k=4)$ .

The equation for the new dilated equation is $y = 4 | x |$ $y=color(red)(4)|x|$ .

Since $k = 4$ $k=4$ this graph will be stretched vertically meaning it will come closer to the y-axis

Use a table of values to find the $y$ $y$ values for both the original and dilated graphs.

$x$ $x$ $1$ $1$ $2$ $2$ $3$ $3$ $4$ $4$

$| x |$ $|x|$ $1$ $1$ $2$ $2$ $3$ $3$ $4$ $4$

$4 | x |$ $4|x|$ $4$ $4$ $8$ $8$ $12$ $12$ $16$ $16$

Sketch the graph of $y = | x |$ $y=|x|$ and $y = 4 | x |$ $y=4|x|$ using the table of values.

Vertical Dilations (Stretch/Shrink)

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Quizzes

$x$ $x$	$\frac{1}{2}$ $1/2$	$1$ $1$	$3$ $3$
$ln x$ $lnx$	$- 0.7$ $-0.7$	$0$ $0$	$1.1$ $1.1$
$\frac{1}{3} ln x$ $1/3lnx$	$- 0.2$ $-0.2$	$0$ $0$	$0.37$ $0.37$

$x$ $x$	$- 2$ $-2$	$- 1$ $-1$	$0$ $0$	$1$ $1$	$2$ $2$
$x^{2}$ $x^2$	$4$ $4$	$1$ $1$	$0$ $0$	$1$ $1$	$4$ $4$
$3 x^{2}$ $3x^2$	$12$ $12$	$3$ $3$	$0$ $0$	$3$ $3$	$12$ $12$

$x$ $x$	$0$ $0$	$1$ $1$	$2$ $2$	$3$ $3$
$e^{x}$ $e^x$	$1$ $1$	$2.7$ $2.7$	$7.4$ $7.4$	$20$ $20$
$\frac{1}{4} e^{x}$ $1/4e^x$	$\frac{1}{4}$ $1/4$	$0.7$ $0.7$	$1.9$ $1.9$	$5$ $5$