Combinations of Transformations: Find Equation 1

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Question 1 of 5

1. Question
Find the equation when $y = x^{2}$ $y=x^2$ is shifted $3$ $3$ units down and $4$ $4$ units to the right.
- 1. $y = {(x - 4)}^{2} - 3$ $y=(x-4)^2-3$
- 2. $y = {(x + 4)}^{2} - 3$ $y=(x+4)^2-3$
- 3. $y = {(x - 3)}^{2} + 4$ $y=(x-3)^2+4$
- 4. $y = {(x + 3)}^{2} + 4$ $y=(x+3)^2+4$
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A standard function form for translations and dilations is $y = a {(x - h)}^{2} + c$ $y=color(red)(a)(x-color(blue)(h))^2+color(purple)(c)$ where:

$a$ $color(red)(a)$ is the vertical dilation (sometimes we use $y = k {(x - h)}^{2} + c$ $y=color(red)(k)(x-color(blue)(h))^2+color(purple)(c)$ )

$h$ $color(blue)(h)$ is the $x$ $x$ -coordinate of the vertex

$c$ $color(purple)(c)$ is the $y$ $y$ -coordinate of the vertex

Vertex $(h, c)$ $(color(blue)(h),color(purple)(c))$

To find the equation when $y = x^{2}$ $y=x^2$ is shifted $3$ $3$ units down and $4$ $4$ units right, remember $y = a {(x - h)}^{2} + c$ $y=color(red)(a)(x-color(blue)(h))^2+color(purple)(c)$ where $a$ $color(red)(a)$ is the vertical dilation, $h$ $color(blue)(h)$ is the $x$ $x$ -coordinate of the vertex, and $c$ $color(purple)(c)$ is the $y$ $y$ -coordinate of the vertex.

The vertex in $y = x^{2}$ $y=x^2$ (which can also be written as $y = {(x - 0)}^{2} + 0)$ $y=(x-0)^2+0)$ is $(0, 0)$ $(0,0)$ . If you shift the vertex $3$ $color(purple)(3)$ units down (downward movement along the $y$ $y$ -axis) and $4$ $color(blue)(4)$ units right (shift to the left along the $x$ $x$ -axis) you will get a new vertex of $(0 + 4, 0 + (- 3)) = (4, - 3)$ $(0+color(blue)(4),0+(color(purple)(-3)))=(color(blue)(4),color(purple)(-3))$

To get the new equation, substitute the new vertex coordinates $(4, - 3)$ $(color(blue)(4),color(purple)(-3))$ into $y = {(x - 0)}^{2} + 0$ $y=(x-0)^2+0$ .
This gives $y = {(x - 4)}^{2} - 3$ $y=(x-color(blue)(4))^2-color(purple)(3)$ which simplifies to $y = {(x - 4)}^{2} - 3$ $y=(x-4)^2-3$ .

$y = {(x - 4)}^{2} - 3$ $y=(x-4)^2-3$
Question 2 of 5

2. Question
Find the equation when ${(x - 1)}^{2} + {(y - 3)}^{2} = 1$ $(x-1)^2 + (y-3)^2 = 1$ is reflected about the $y$ $y$ axis and then shifted down $2$ $2$ units.
- 1. ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ $(x-1)^2 + (y-1)^2 = 1$
- 2. ${(x + 1)}^{2} + {(y - 1)}^{2} = 1$ $(x+1)^2 + (y-1)^2 = 1$
- 3. ${(x - 1)}^{2} + {(y + 1)}^{2} = 1$ $(x-1)^2 + (y+1)^2 = 1$
- 4. ${(x + 1)}^{2} + {(y + 1)}^{2} = 1$ $(x+1)^2 + (y+1)^2 = 1$
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Reflections about the $y$ $y$ -axis means that we have to replace $x \to - x$ $x\rightarrow-x$ .

A standard circle equation is in the form: ${(x - h)}^{2} + {(y - c)}^{2} = r^{2}$ $(x-color(blue)(h))^2 + (y-color(purple)(c))^2 =r^2$

$h$ $color(blue)(h)$ is the $x$ $x$ -coordinate for the centre

$c$ $color(purple)(c)$ is the $y$ $y$ -coordinate for the centre

First we find the equation when ${(x - 1)}^{2} + {(y - 3)}^{2} = 1$ $(x-1)^2 + (y-3)^2 = 1$ is reflected about the $y$ $y$ -axis, we replace the $x$ $x$ by $- x$ $-x$ . We get ${(- x - 1)}^{2} + {(y - 3)}^{2} = 1$ $(-x-1)^2 + (y-3)^2 = 1$ or simply ${(x + 1)}^{2} + {(y - 3)}^{2} = 1$ $(x+1)^2 + (y-3)^2 = 1$ .

Remember ${(x - h)}^{2} + {(y - c)}^{2} = r^{2}$ $(x-color(blue)(h))^2 + (y-color(purple)(c))^2 =r^2$ where $h$ $color(blue)(h)$ is the $x$ $x$ -coordinate of the centre, and $c$ $color(purple)(c)$ is the $y$ $y$ -coordinate of the centre.

The center for ${(x + 1)}^{2} + {(y - 3)}^{2} = 1$ $(x+1)^2 + (y-3)^2 = 1$ is $(- 1, 3)$ $(-1,3)$ .

Next we will shift this circle $2$ $2$ units down along the $y -$ $y-$ axis. Since the formula indicates ${(y - c)}^{2}$ $(y-c)^2$ it becomes ${(y - - c)}^{2} \to {(y + c)}^{2} .$ $(y- -c)^2 \rightarrow (y+c)^2.$

Notice it is $+ c$ $color(purple)(+c)$ even though we are going downwards. So as a rule of thumb for this formula: ${(x - h)}^{2} + {(y - c)}^{2} = r^{2}$ $(x-color(blue)(h))^2 + (y-color(purple)(c))^2 =r^2$ anytime $- c$ $color(purple)(-c)$ or $- h$ $color(blue)(-h)$ is inside the brackets we move in the opposite direction.

This gives ${(x + 1)}^{2} + {(y - 1)}^{2} = 1 .$ $(x+1)^2+(y-1)^2=1.$
The centre is $(- 1, 1)$ $(-1,1)$ .

${(x + 1)}^{2} + {(y - 1)}^{2} = 1$ $(x+1)^2 + (y-1)^2 = 1$
Question 3 of 5

3. Question
Find the equation when $y = e^{x}$ $y=e^x$ is vertically dilated by a factor of $4$ $4$ and shifted $5$ $5$ units to the left.
- 1. $y = 4 e^{x + 5}$ $y=4e^(x+5)$
- 2. $y = (\frac{1}{4}) e^{x + 5}$ $y=(1/4)e^(x+5)$
- 3. $y = 4 e^{x - 5}$ $y=4e^(x-5)$
- 4. $y = e^{4 x + 5}$ $y=e^(4x+5)$
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A standard exponential function form for translations and dilations is

$k$ $color(red)(k)$ is the vertical dilation $y = k e^{(x + h)}$ $y=color(red)(k)e^((x+color(blue)(h)))$

$h$ $color(blue)(h)$ is the $x$ $x$ -coordinate for the horizontal translation

To find the equation when $y = e^{x}$ $y=e^x$ is vertically dilated by a factor of $4$ $4$ and shifted $5$ $5$ units left, remember $y = k e^{(x + h)}$ $y=color(red)(k)e^((x+color(blue)(h)))$ where $k$ $color(red)(k)$ is the vertical dilation and $h$ $color(blue)(h)$ is the horizontal translation.

The equation $y = e^{x}$ $y=e^x$ when vertically dilated by a factor of $4$ $4$ becomes $y = 4 e^{x}$ $y=4e^x$ . Then applying the translation of $5$ $5$ units to the left
(left means $+ 5$ $+5$ ), $y = 4 e^{x}$ $y=4e^x$ becomes $y = 4 e^{x + 5}$ $y=4e^(x+5)$ .

$y = 4 e^{x + 5}$ $y=4e^(x+5)$

Question 4 of 5

Find the equation when $y = 2 x^{2} - 3 x$ $y=2x^2 -3x$ is shifted $1$ $1$ unit left and then reflected about the $y$ $y$ axis.

1. $y = 2 x^{2} - x - 1$ $y=2x^2-x-1$
2. $y = 2 x^{2} + x - 1$ $y=2x^2 +x-1$
3. $y = 2 x^{2} + 2 x + 1$ $y=2x^2 +2x+1$
4. $y = x^{2} - 2 x - 1$ $y=x^2-2x-1$

Question 5 of 5

Find the transformed version of $y = x^{3}$ $y=x^3$ when you have a horizontal dilation factor of $\frac{1}{2}$ $1/2$ , a horizontal translation of $3$ $3$ units left, and a vertical dilation of $4$ $4$ are applied.

1. $y = 32 {(2 x + 3)}^{3}$ $y=32(2x+3)^3$
2. $y = 4 {(x + 3)}^{3}$ $y=4(x+3)^3$
3. $y = 32 {(x - 3)}^{3}$ $y=32(x-3)^3$
4. $y = 32 {(x + 3)}^{3}$ $y=32(x+3)^3$

$y$ $y$	$=$ $=$	$2 {(x + 1)}^{2} - 3 (x + 1)$ $2(x color(blue)(+1))^2-3(x color(blue)(+1))$
$y$ $y$	$=$ $=$	$2 (x^{2} + 2 x + 1) - 3 x - 3$ $2(x^2+2x+1)-3x-3$
$y$ $y$	$=$ $=$	$2 x^{2} + 4 x + 2 - 3 x - 3$ $2x^2+4x+2-3x-3$
$y$ $y$	$=$ $=$	$2 x^{2} + x - 1$ $2x^2+x-1$

$y =$ $y=$	${(\frac{x}{\frac{1}{2}})}^{3}$ $(x/color(blue)(1/2))^3$	Apply the horizontal dilation factor of $\frac{1}{2}$ $color(blue)(1/2)$ . Remember $\frac{x}{factor}$ $x/color(blue)(text{factor})$ .
$=$ $=$	${(2 x)}^{3}$ $(2x)^3$	Simplify
$=$ $=$	${(2 (x + 3))}^{3}$ $(2(xcolor(red)(+3)))^3$	Apply the horizontal translation of $3$ $color(red)(3)$ units left. Use $y = f (x + 3)$ $y=f(xcolor(red)(+3))$ .
$=$ $=$	${(2 (x + 3))}^{3}$ $(2(x+3))^3$	Simplify

$y =$ $y=$	$4 {(2 (x + 3))}^{3}$ $color(blue)(4)(2(x+3))^3$	Apply the vertical dilation of $k = 4$ $color(blue)(k=4)$ . Use $k f (x)$ $color(blue)(k)f(x)$ .
$=$ $=$	$4 {(2 (x + 3))}^{3}$ $4(2(x+3))^3$	Simplify
$=$ $=$	$32 {(x + 3)}^{3}$ $32(x+3)^3$

Combinations of Transformations: Find Equation 1

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