Answer with explanation:
The given function is
y=f(x)=x³ -5
Function with Translation
[tex]1.\rightarrow f(\frac{x}{5})=[\frac{x}{5}]^3-5\\\\y_{1}=\frac{x^3}{125}-5\\\\2..\rightarrow f(x-5)=(x-5)^3-5\\\\y_{2}=x^3-5^3-3x^2\times 5+3 x \times 5^2-5\\\\y_{2}=x^3-15 x^2+75 x-130 \\\\3.\rightarrow f(x)+10=x^3-5+10\\\\y_{3}=x^3+5\\\\4.\rightarrow f(5 x-2)=(5 x-2)^3-5\\\\y_{4}=(5 x)^3-(2)^3-3 \times (5 x)^2 \times 2+3 \times (5 x) \times (2)^2-5\\\\y_{4}=125 x^3-8-150 x^2+60 x-5\\\\y_{4}=125 x^3-150 x^2+60 x-13[/tex]
Answer:
The graph of function shown below.
Step-by-step explanation:
The given function is
[tex]y=f(x)=x^3-5[/tex]
The graph of this function is the graph of parent function [tex]g(x)=x^3[/tex] after shifting 5 units down.
The translation is defined as
[tex]g(x)=f(kx+a)^2+b[/tex] .... (2)
Where, k is stretch factor, a is horizontal shift and b is vertical shift.
If 0<k<1, then the graph stretch horizontally by factor 1/k and if k>1, then the graph compressed horizontally by factor 1/k.
If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.
If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.
The given functions are :
1. [tex]y=f(\frac{x}{5})[/tex]
Here, k=1/5. So the graph of f(x) stretch horizontally by factor 5.
2. [tex]y=f(x-5)[/tex]
Here, a=-5, So, the graph of f(x) shifts 5 units right.
3. [tex]y=f(x)+10[/tex]
Here, b=10, So, the graph of f(x) shifts 10 units up.
4. [tex]y=f(x)+10[/tex]
Here, k=5 and a=-2, So, the graph of f(x) compressed horizontally by factor 1/5 and shifts 2 units right.
The graph of function shown below.
