Which equation defines the graph of y=x^3 after it is shifted vertically 5 units down and horizontally left 4 units
a) y=(x-4)^3 - 5
b) y=(x+5)^3 - 4
c) y=(x+5)^3 + 4
d) y=(x+4)^3 - 5

Option D

The function shifted vertically 5 units down and horizontally left 4 units is,

Step-by-step explanation

The function is given by, y=

To move a function shift vertically 5 units down

We know that moving the function down, you subtract outside the function i.e, ; moved down c units.

Therefore, the function f(x) vertically shift 5 units down by  

Now, to move a function horizontally 4 units left.

To shift the function left add inside the function's argument i.e, gives f(x) shifted a units to the left.

so, the function f(x) horizontally left 4 units i.e, f(x+4)=

Now, the graph of the function after it is shifted vertically 5 units down and horizontally left 4 units is,

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

answer: 0.64

step-by-step explanation: