Anyone add my sna-p need help with my geometry, it's callmelako
1 answer:
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Answer:
# The transformation of f(x) to be g(x) is
- Reflected across the x-axis
- Translated 7 units to the left
- Translated 6 units up
# The transformation of f(x) to be g(x) is
- Reflected across the y-axis
- Translated 7 units down
Step-by-step explanation:
* Lets revise some transformation rules
* Lets talk about the transformation
- If the function f(x) reflected across the x-axis, then the new
function g(x) = - f(x)
- If the function f(x) reflected across the y-axis, then the new
function g(x) = f(-x)
- If the function f(x) translated horizontally to the right
by h units, then the new function g(x) = f(x - h)
- If the function f(x) translated horizontally to the left
by h units, then the new function g(x) = f(x + h)
- If the function f(x) translated vertically up
by k units, then the new function g(x) = f(x) + k
- If the function f(x) translated vertically down
by k units, then the new function g(x) = f(x) – k
* Lets solve the problems
# f(x) = 2^x ⇒ g(x) = -(2)^(x + 7) + 6
- There is a -ve sign in-front of the 2
∵ f(x) will be -f(x)
∴ f(x) will reflect across the x-axis ⇒ (1)
- The power x becomes x + 7
∵ -f(x) will be -f(x + 7)
∴ -f(x) will translate 7 units to the left ⇒ (2)
- after that -f(x + 7) add by 6
∴ -f(x + 7) will translate 6 units up ⇒ (3)
- From (1) , (2) , (3)
∴ The transformation of f(x) to be g(x) is
- Reflected across the x-axis
- Translated 7 units to the left
- Translated 6 units up
# f(x) = ㏒(x) ⇒ g(x) = ㏒(-x) - 7
- The (x) will be (-x)
∵ f(x) will be f(-x)
∴ f(x) will reflect across the y-axis ⇒ (1)
- after that f(-x) subtracted by 7
∴ f(-x) will translate 7 units down ⇒ (2)
- From (1) , (2)
∴ The transformation of f(x) to be g(x) is
- Reflected across the y-axis
- Translated 7 units down
* For more understand look to the attached graphs
# First function:
- f(x) is the red
- g(x) is the blue
# Second function:
- f(x) is the black
- g(x) is the green
To solve
replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)
so
I'm not sure if it is
![f(x)=\sqrt[3]{\frac{x}{7}}-9](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx%7D%7B7%7D%7D-9)
or
![f(x)=\sqrt[3]{\frac{x}{7}-9}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx%7D%7B7%7D-9%7D)
first one
replace f(x) with y
![y=\sqrt[3]{\frac{x}{7}}-9](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx%7D%7B7%7D%7D-9)
switch x and y
![x=\sqrt[3]{\frac{y}{7}}-9](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7By%7D%7B7%7D%7D-9)
solve for y
![x+9=\sqrt[3]{\frac{y}{7}}](https://tex.z-dn.net/?f=x%2B9%3D%5Csqrt%5B3%5D%7B%5Cfrac%7By%7D%7B7%7D%7D)
replace y with f⁻¹(x)
2nd one
replce f(x) with y
![y=\sqrt[3]{\frac{x}{7}-9}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx%7D%7B7%7D-9%7D)
switch x and y
![x=\sqrt[3]{\frac{y}{7}-9}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7By%7D%7B7%7D-9%7D)
solve for y
replace y with f⁻¹(x)
if you meant then the inverse is
if you meant then the inverse is
replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)
so
I'm not sure if it is
first one
replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)
2nd one
replce f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)
if you meant
if you meant