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Year 12>
Transformations of Functions>
Write the Equation from Translations>
Write the Equation from TranslationsWrite the Equation from Translations
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Question 1 of 6
1. Question
Find the equation of the function when `y=x^2` is shifted (translated) to the right by `2` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) of a function have the form `y=(x- color(royalblue)(h))^2+ color(forestgreen)(c)` where the point `( color(royalblue)(h), color(forestgreen)(c))` is the vertex on the function. Horizontal translations have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift LeftTo obtain the equation of the function by using `y=x^2`, shift the graph to the right by `2` units. Remember squared functions have the form `y=(x- color(royalblue)(h))^2+ color(forestgreen)(c)`.This shift is a horizontal translation (`h`) where `h=-2` inside the brackets.`y=` `x^2` Write the new equation using `y=x^2` and `h=-2`. Remember horizontal translations have the opposite sign as the direction of their shift and squared functions have the form `y=(x-h)^2+c`. `y=` `(x-color(blue)(2))^2` `y=(x-2)^2` -
Question 2 of 6
2. Question
Find the equation of the function when `y=logx` is shifted (translated) to the left by `5` units and down by `2` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) of a function here is in the form of `y=log(x-color(blue)(h))+color(forestgreen)(c)` where the point `( color(royalblue)(h), color(forestgreen)(c))` is on the function. Horizontal translations have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the equation of the function by using `y=logx`, first shift the graph to the left by `5` units.This shift is a horizontal translation left, where `color(blue)(h=+5)` inside the brackets.Next, shift the graph down by `color(forestgreen)(-2)` units.This shift is a vertical translation (`c`) where `color(forestgreen)(c=-2)`.`y=log(x+5)-2` -
Question 3 of 6
3. Question
Find the equation of the function when `x^2 + y^2 =9` is shifted (translated) to the right by `2` units and up by `3` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) for a circle have the form `(x-color(blue)(h))^2 + (y-color(forestgreen)(c))^2=r^2` where the point `( color(royalblue)(h), color(forestgreen)(c))` is the center. Horizontal and vertical translations for circles have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(uarr)` Shift Up`(+c) \ bb(darr)` Shift DownTo obtain the equation of the function by using `x^2 + y^2=r^2`, first shift the graph to the right by `2` units. Remember circle equations have the form `(x-color(blue)(h))^2 + (y-color(forestgreen)(c))^2=r^2`.This shift is a horizontal translation (`h`) (`2` units to the right) where `color(blue)(h=-2)` inside the brackets.Next, shift the graph vertically up by `3` units where `color(forestgreen)(c=-3)` inside the brackets.`x^2 + y^2=` `9` Since it’s in the form `(x-color(blue)(h))^2 + (y-color(forestgreen)(c))^2=r^2`, then `(x-color(blue)(2))^2 + (y-color(forestgreen)(3))^2=9`.
Remember with circles, that the vertical and horizontal translations have the opposite sign since they are inside the brackets.`(x-color(blue)(2))^2 + (y-color(forestgreen)(3))^2=` `9` `(x-2)^2 + (y-3)^2=9` -
Question 4 of 6
4. Question
Find the equation of the function when `y=sqrt(x)` is shifted (translated) to the left by `2` units and up by `4` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) for a square root function has the form `y=sqrt(x-color(blue)(h))+color(forestgreen)(c)` where the point `( color(royalblue)(h), color(forestgreen)(c))` is the vertex for this function. Horizontal translations have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the equation of the function by using `y=sqrt(x)`, first shift the graph to the left by `2` units. Remember square root functions have the form `y=sqrt(x-color(blue)(h))+color(forestgreen)(c)`.This shift is a horizontal translation (`h`) where `color(blue)(h=+2)` inside the square root.Next, shift the graph up by `4` units `color(forestgreen)((c=4))``y=sqrt(x+2)+4` -
Question 5 of 6
5. Question
Find the equation of the function when `y=1/x` is shifted (translated) to the left by `3` units and up by `2` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) for a hyperbolic function is in the form `y=1/(x-color(blue)(h))+color(forestgreen)(c)` where `( color(royalblue)(h), color(forestgreen)(c))` is the origin for this function. Horizontal translations have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the equation of the function by using `y=1/x`, first shift the graph to the left by `3` units. Remember hyperbolas have the form `y=1/(x-color(blue)(h))+color(forestgreen)(c)`.Firstly, we shift a horizontal translation (left), where `color(blue)(h=+3)` in the denominator.Next, shift a vertical translation up by `2` units `color(forestgreen)((c=2))`.`y=` `1/x` Write the new equation using `y=1/x`, `color(forestgreen)(c=2)`, and `color(blue)(h=+3)`. Remember horizontal translations have the opposite sign as the direction of their shift and hyperbolas have the form `y=1/(x-color(blue)(h))+color(forestgreen)(c)`. `y=` `1/(x+color(blue)(3))+color(forestgreen)(2)` `y=1/(x+3)+2` -
Question 6 of 6
6. Question
Find the equation of the function when `y=2^(x-2)` is shifted (translated) to the left by `3` units and up by `4` units.
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Horizontal translations (`\color(royalblue)(h)`) and vertical translations (`\color(forestgreen)(c)`) for an exponential function have the form `y=2^(x-color(blue)(h)) + color(forestgreen)(c)` where the point `( color(royalblue)(h), color(forestgreen)(c))` is on the function. Horizontal translations have the opposite sign as the direction of their shift.`(-h) \ bb(rarr)` Shift Right`(+h) \ bb(larr)` Shift Left`(-c) \ bb(darr)` Shift Down`(+c) \ bb(uarr)` Shift UpTo obtain the equation of the function by using `y=2^(x-2)`, first shift the graph to the left by `3` units. Remember exponential functions have the form `y=2^(x-color(blue)(h)) + color(forestgreen)(c)`.This shift is a horizontal translation (left), where `color(blue)(h=+3)`.Next, is a vertical translation up by `4` units `color(forestgreen)((c=4))`.`y=` `2^(x-2)` Write the new equation using `y=2^(x-2)`, `color(forestgreen)(c=4)`, and `color(blue)(h=+3)`. Remember horizontal translations have the opposite sign as the direction of their shift and exponential functions have the form `y=2^(x-color(blue)(h)) + color(forestgreen)(c)`. `y=` `2^(x-2+color(blue)(3))+color(forestgreen)(4)` Simplifying the exponent. `y=` `2^(x+1)+4` `y=2^(x+1)+4`
Quizzes
- Vertical Translations 1
- Vertical Translations 2
- Vertical Translations from a Point
- Horizontal Translations 1
- Horizontal Translations 2
- Horizontal Translations from a Point
- Horizontal Translations from a Graph
- Horizontal and Vertical Translations from a Graph
- Sketch a Graph using Translations
- Write the Equation from a Graph
- Write the Equation from Translations
- Vertical Dilations
- Horizontal Dilations 1
- Horizontal Dilations 2
- Horizontal Dilations – Scale Factor
- Horizontal and Vertical Dilations 1
- Horizontal and Vertical Dilations 2
- Horizontal and Vertical Dilations 3
- Graphing Reflections 1
- Graphing Reflections 2
- Reflection with Rotation
- Combinations of Transformations: Order
- Combinations of Transformations: Coordinates
- Combinations of Transformations: Find Equation 1
- Combinations of Transformations: Find Equation 2
- Combinations of Transformations: Find Equation 3