Respuesta :

naǫ
[tex]f(x) \xrightarrow{\hbox{1 unit to the left}} f(x+1) \\ \\ f(x)=x^3+x^2-2x+1 \\ \\ f(x+1)=(x+1)^3+(x+1)^2-2(x+1)+1= \\ =x^3+3x^2+3x+1+x^2+2x+1-2x-2+1= \\ =x^3+4x^2+3x+1[/tex]

The resulting function is:
[tex]\boxed{y=x^3+4x^2+3x+1}[/tex]
carlosego

For this case we have the following function:

[tex] f (x) = x ^ 3 + x ^ 2 - 2x + 1
[/tex]

We apply the following function transformation:

Horizontal translations:

Suppose that h> 0

To graph y = f (x + h), move the graph of h units to the left.

We have then for h = 1:

[tex] f (x + 1) = (x + 1) ^ 3 + (x + 1) ^ 2 - 2 (x + 1) + 1
[/tex]

Rewriting we have:

[tex] f (x + 1) = (x ^ 3 + 3x ^ 2 + 3x + 1) + (x ^ 2 + 2x + 1) + (-2x - 2) + 1
[/tex]

Rewriting we have:

[tex] f (x + 1) = x ^ 3 + 4x ^ 2 + 3x + 1 [/tex]

Answer:

The resulting function when f (x) is shifted to the left 1 unit is:

[tex] f (x + 1) = x ^ 3 + 4x ^ 2 + 3x + 1 [/tex]

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