a.

There is no vertical stretch, so $\left|a\right|=1$.

There is a horizontal stretch about the $y$-axis by a factor of $2$, so $\left|b\right|=\frac{1}{2}$.

There is a reflection in the $x$-axis, so all of the $y$-values change to the opposite sign.

Because the original function was $y=x^2$ or $f\left(x\right)=x^2$, the $x$ is replaced with $\frac{1}{2}x$ and a negative sign is placed in front.

b.

There is a horizontal stretch about the $y$-axis by a factor of $\frac{1}{4}$, so $\left|b\right|=4$.


There is a reflection in the $y$-axis, so all of the $x$-values change to the opposite sign.


There is a vertical stretch about the $x$-axis by a factor of $\frac{1}{4}$ and there is no vertical reflection, so $a=\frac{1}{4}$.


Because the original function was $y=x^2$ or $f\left(x\right)=x^2$. The $x$ is replaced with $-4x$ and $\frac{1}{4}$ is placed in front.

a.

First, check for stretches and reflections. Then, look for translations.

There were no vertical or horizontal stretches because the shape of the graph has not been changed. The values of $\left|a\right|$ and $\left|b\right|$ are $1$.


There has been a reflection in the $y$-axis because the orientation of the graph has changed. A horizontal portion of the graph of the function $y=f\left(x\right)$ lies in quadrant II, and there is a corresponding horizontal section of the transformed graph in quadrant I. The equation of a reflection in the $y$-axis is $y=f\left(-x\right)$. Therefore, the equation of the transformed graph is


We can determine if a translation has taken place by locating key points on the graph of $f\left(x\right)$ and the image points on the graph of the transformed function.


If there had been only a reflection, the point $\left(-6\mathrm{,}4\right)$ would map to the image point $\left(6\mathrm{,}4\right)$. Instead, it has mapped to the point $\left(4\mathrm{,}2\right)$, which means there was a horizontal shift to the left $2$ units $\left(6-4=2\right)$ and a vertical shift downward $2$ units $\left(4-2=2\right)$.

Checking the next key point gives the same result. If there had been only a reflection, the point $\left(-2\mathrm{,}4\right)$ would map to the image point $\left(2\mathrm{,}4\right)$. Instead, it has mapped to the point $\left(0\mathrm{,}2\right)$, which means there was a horizontal shift $2$ units left and a vertical shift $2$ units down. A check of the other two key points will yield the same result.

Therefore, $h=-2$ and $k=-2$. The equation of the transformed function is $y=f\left(-\left(x+2\right)\right)-2$.

b.

First, check for stretches and reflections. Then, look for translations. A stretch has occurred because the shape of the transformed graph is not the same as the graph of $y=f\left(x\right)$. By looking at the domain and range of the transformed graph, we can determine if a horizontal or vertical stretch has occurred.

The range has changed, but spans the same vertical distance from minimum to maximum as the range of $y=f\left(x\right)$. As such, the graph of the transformed function has not been vertically stretched, and the value of $a=1$.

The domain has also changed, but it also spans a different horizontal distance from left to right compared to the domain of $y=f\left(x\right)$. As such, the graph of the transformed function has been horizontally stretched.


Determine the horizontal stretch factor by comparing the distance between key points on the graphs. Consider the horizontal section from $\left(-6\mathrm{,}4\right)$ to $\left(-2\mathrm{,}4\right)$ on the graph of the original function. It is $4$ units long.

On the graph of the transformed function, the corresponding horizontal section is from $\left(-4\mathrm{,}0\right)$ to $\left(-2\mathrm{,}0\right)$, which is $2$ units long.

The function underwent a horizontal stretch about the $y$-axis by a factor of $\frac{1}{2}$, so $b=2$.

Therefore, $a=1$ and $b=2$. The equation of the transformed function is


Because the orientation of the graph of the transformed function is the same as the orientation of the graph of $y=f\left(x\right)$, there have been no reflections.

We can determine if a translation has taken place by locating key points on the graph of $f\left(x\right)$ and the image points on the graph of the transformed function.


Each $y$-value is $4$ less on the transformed function. Therefore, the points have been translated down $4$ units, and $k=-4$.

Use mapping again to check for a horizontal translation. If there had only been a horizontal stretch and a vertical translation, the point $\left(-6\mathrm{,}4\right)$ would map to the image point $\left(-3\mathrm{,}0\right)$. Instead, it has mapped to the point $\left(-4\mathrm{,}0\right)$, which means there was a horizontal translation $1$ unit left, and $h=-1$. The result is the same if we check the other key points.

Therefore, $h=-1$ and $k=-4$. The equation of the transformed function is

a.

Compare $y=2f\left(x-3\right)+4$ to $y=af\left(b\left(x-h\right)\right)+k$.

$a=2$ indicates a vertical stretch about the $x$-axis by a factor of $2$. All the $y$-values will be multiplied by $2$.


$b=1$ indicates there is no horizontal stretch.


$h=3$ indicates a horizontal translation $3$ units right. All the $x$-values will have $3$ added to them.


$k=4$ indicates a vertical translation $4$ units up. All the $y$-values will have $4$ added to them.

c.

Compare $y=-\frac{1}{4}f\left(-\left(x+2\right)\right)$ to $y=af\left(b\left(x-h\right)\right)+k$.

$a=-\frac{1}{4}$ indicates a reflection in the $x$-axis. All the $y$-values will change to the opposite sign. $\left(x\mathrm{,}y\right)\to \left(x\mathrm{,}-y\right)$.

It also indicates a vertical stretch about the $x$-axis by a factor of $\frac{1}{4}$. All the $y$-values will be multiplied by $\frac{1}{4}$. $\left(x\mathrm{,}y\right)\to \left(x\mathrm{,}-\frac{1}{4}y\right)$.

$b=-1$ indicates a reflection in the $y$-axis. All the $x$-values will change to the opposite sign.


$h=-2$ indicates a horizontal translation $2$ units left. All the $x$-values will have $2$ subtracted from them.


$k=0$ indicates there is no vertical translation.

f.

First write $3y-6=f\left(-2x+12\right)$ in the form $y=af\left(b\left(x-h\right)\right)+k$.


Now, the values of $a\mathrm{,}b\mathrm{,}h\mathrm{,}$ and $k$ can be read.

$a=\frac{1}{3}$ indicates a vertical stretch about the $x$-axis by a factor of $\frac{1}{3}$. All the $y$-values will be multiplied by $\frac{1}{3}$.


$b=-2$ indicates a reflection in the $y$-axis. All the $x$-values will change to the opposite sign $\left(x\mathrm{,}y\right)\to \left(-x\mathrm{,}\frac{1}{3}y\right)$.

It also indicates a horizontal stretch about the $y$-axis by a factor of $\frac{1}{2}$. All the $x$-values will be multiplied by $\frac{1}{2}$.


$h=6$ indicates a horizontal translation $6$ units right. All the $x$-values will have $6$ added to them.


$k=2$ indicates a vertical translation $2$ units up. All the $y$-values will have $2$ added to them.

a.

A horizontal stretch about the $y$-axis by a factor of $2$ means all the $x$-values are multiplied by $2$.


A reflection in the $x$-axis means all the $y$-values change to the opposite sign.


A translation $4$ units up means $4$ is added to all the $y$-values.


A translation $3$ units left means $3$ is subtracted from all the $x$-values.


So,  $A\left(-4\mathrm{,}6\right)$ maps to $\left(2\left(-4\right)-3\mathrm{,}-\left(6\right)+4\right)=\left(-11\mathrm{,}-2\right)$.

Follow the same process for the other points.

b.

A horizontal stretch about the $y$-axis by a factor of $2$ means $b=\frac{1}{2}$.

A reflection in the $x$-axis means $a$ is negative, and $\left|a\right|=1$.

A translation $4$ units up means $k=4$.

A translation $3$ units left means $h=-3$.

Substitute the values of $a\mathrm{,}b\mathrm{,}h\mathrm{,}$ and $k$ into $y=af\left(b\left(x-h\right)\right)+k$ to get $y=-f\left(\frac{1}{2}\left(x+3\right)\right)+4$.