Question

1. graph the function $f(x) = -(x + 2)^3$ and find the following features.

a. point of inflection: ______________
b. zero(s): ____________
c. domain (interval form): ______________
d. range (interval form): ______________
e. transformations:__________________________
f. end behavior: on the left, the graph ______________
on the right, the graph ______________

1. graph the function $f(x) = (x - 2)^3 + 3$ and find the following features.

a. domain (inequality form): ____________________
b. range (inequality form): ____________________
c. point of inflection: ____________
d. transformations:__________________________
e. end behavior: on the left, the graph ______________
on the right, the graph ______________

1. if $f(x) = x^3$, graph and describe the effects if it is transformed as follows: $-f(x + 4) - 2$.

a. domain(set builder): ____________
b. range(set builder): ____________
c. point of inflection: ____________
d. transformations:__________________________
e. end behavior: on the left, the graph ______________
on the right, the graph ______________

1. write the equation of a cubic function that has been translated down three units, right one unit and is vertically compressed by a factor of $\frac{1}{2}$.

equation: __________________________

Answer:

Problem 1: Function \( f(x) = -(x + 2)^3 \)

a. Point of Inflection

The parent function of a cubic function \( y = x^3 \) has a point of inflection at \( (0, 0) \). For \( f(x) = -(x + 2)^3 \), the graph is a transformation of \( y = x^3 \) (reflected over the x - axis and shifted left 2 units). The point of inflection of a cubic function \( y = a(x - h)^3 + k \) is at \( (h, k) \). Here, \( h=-2 \) and \( k = 0 \), so the point of inflection is \( (-2, 0) \).

b. Zero(s)

To find the zeros, set \( f(x)=0 \):
\( 0=-(x + 2)^3 \)
Multiply both sides by - 1: \( 0=(x + 2)^3 \)
Take the cube root of both sides: \( x+2 = 0 \), so \( x=-2 \). The zero is \( x=-2 \).

c. Domain (Interval form)

Cubic functions are polynomials, and polynomials are defined for all real numbers. So the domain is \( (-\infty, \infty) \).

d. Range (Interval form)

The range of a cubic function (since it is a polynomial of odd degree) is all real numbers. So the range is \( (-\infty, \infty) \).

e. Transformations

The parent function is \( y = x^3 \). The function \( f(x)=-(x + 2)^3 \) is obtained by:

• Horizontal shift: shifting the graph of \( y = x^3 \) 2 units to the left (because of \( x+2 \) in the argument).
• Reflection: reflecting the graph over the x - axis (because of the negative sign in front of \( (x + 2)^3 \)).

f. End Behavior

For a cubic function of the form \( y = ax^3+bx^2+cx + d \), when \( a<0 \) (in our case, the leading coefficient of \( f(x)=-(x + 2)^3=-x^3-6x^2 - 12x - 8 \) is - 1, which is negative):

• As \( x

ightarrow-\infty \) (on the left), \( y
ightarrow\infty \) (because \( \lim_{x
ightarrow-\infty}-x^3=\infty \)).

• As \( x

ightarrow\infty \) (on the right), \( y
ightarrow-\infty \) (because \( \lim_{x
ightarrow\infty}-x^3=-\infty \)).

Problem 2: Function \( f(x)=(x - 2)^3+3 \)

a. Domain (Inequality form)

Cubic functions are polynomials, so they are defined for all real numbers. In inequality form, the domain is \( x\in\mathbb{R} \) (or \( -\infty

b. Range (Inequality form)

Cubic functions have a range of all real numbers. So the range is \( y\in\mathbb{R} \) (or \( -\infty

c. Point of Inflection

For a cubic function \( y = a(x - h)^3 + k \), the point of inflection is at \( (h, k) \). Here, \( h = 2 \) and \( k=3 \), so the point of inflection is \( (2, 3) \).

d. Transformations

The parent function is \( y = x^3 \). The function \( f(x)=(x - 2)^3+3 \) is obtained by:

• Horizontal shift: shifting the graph of \( y = x^3 \) 2 units to the right (because of \( x - 2 \) in the argument).
• Vertical shift: shifting the graph 3 units up (because of the + 3 at the end).

e. End Behavior

The leading coefficient of \( f(x)=(x - 2)^3+3=x^3-6x^2 + 12x - 8 + 3=x^3-6x^2+12x - 5 \) is 1 (positive). For a cubic function with a positive leading coefficient:

• As \( x

ightarrow-\infty \) (on the left), \( y
ightarrow-\infty \) (because \( \lim_{x
ightarrow-\infty}x^3=-\infty \)).

• As \( x

ightarrow\infty \) (on the right), \( y
ightarrow\infty \) (because \( \lim_{x
ightarrow\infty}x^3=\infty \)).

Problem 3: Transformation of \( f(x)=x^3 \) to \( -f(x + 4)-2=- (x + 4)^3-2 \)

a. Domain (Set Builder)

The function \( y=- (x + 4)^3-2 \) is a polynomial, so it is defined for all real numbers. In set - builder notation, the domain is \( \{x|x\in\mathbb{R}\} \).

b. Range (Set Builder)

Since it is a cubic function (a polynomial of odd degree), the range is all real numbers. In set - builder notation, the range is \( \{y|y\in\mathbb{R}\} \).

c. Point of Inflection

For the function \( y=- (x + 4)^3-2 \), which is in the form \( y = a(x - h)^3 + k \) with \( a=-1 \), \( h=-4 \), and \( k=-2 \), the point of inflection is \( (-4, -2) \).

d. Transformations

The parent function is \( y = x^3 \). The function \( y=- (x + 4)^3-2 \) is obtained by:

• Horizontal shift: shifting the graph of \( y = x^3 \) 4 units to the left (because of \( x + 4 \) in the argument).
• Reflection: reflecting the graph over the x - axis (because of the negative sign in front of \( (x + 4)^3 \)).
• Vertical shift: shifting the graph 2 units down (because of the - 2 at the end).

e. End Behavior

The leading coefficient of \( y=- (x + 4)^3-2=-x^3-12x^2-48x - 64 - 2=-x^3-12x^2-48x - 66 \) is - 1 (negative).

• As \( x

ightarrow-\infty \) (on the left), \( y
ightarrow\infty \) (since \( \lim_{x
ightarrow-\infty}-x^3=\infty \)).

• As \( x

ightarrow\infty \) (on the right), \( y
ightarrow-\infty \) (since \( \lim_{x
ightarrow\infty}-x^3=-\infty \)).

Problem 4: Equation of the Cubic Function

The parent cubic function is \( y = x^3 \).

• Vertical compression by a factor of \( \frac{1}{2} \): multiply the function by \( \frac{1}{2} \), so we get \( y=\frac{1}{2}x^3 \).
• Translation right 1 unit: replace \( x \) with \( x - 1 \), so we get \( y=\frac{1}{2}(x - 1)^3 \).
• Translation down 3 units: subtract 3 from the function, so we get \( y=\frac{1}{2}(x - 1)^3-3 \).

Final Answers

Problem 1

a. Point of Inflection: \( \boldsymbol{(-2, 0)} \)
b. Zero(s): \( \boldsymbol{x=-2} \)
c. Domain: \( \boldsymbol{(-\infty, \infty)} \)
d. Range: \( \boldsymbol{(-\infty, \infty)} \)
e. Transformations: Shift left 2, reflect over x - axis
f. End Behavior: On the left, the graph \( \boldsymbol{
ightarrow\infty} \); On the right, the graph \( \boldsymbol{
ightarrow-\infty} \)

Problem 2

a. Domain: \( \boldsymbol{-\inftyb. Range: \( \boldsymbol{-\inftyc. Point of Inflection: \( \boldsymbol{(2, 3)} \)
d. Transformations: Shift right 2, shift up 3
e. End Behavior: On the left, the graph \( \boldsymbol{
ightarrow-\infty} \); On the right, the graph \( \boldsymbol{
ightarrow\infty} \)

Problem 3

a. Domain: \( \boldsymbol{\{x|x\in\mathbb{R}\}} \)
b. Range: \( \boldsymbol{\{y|y\in\mathbb{R}\}} \)
c. Point of Inflection: \( \boldsymbol{(-4, -2)} \)
d. Transformations: Shift left 4, reflect over x - axis, shift down 2
e. End Behavior: On the left, the graph \( \boldsymbol{
ightarrow\infty} \); On the right, the graph \( \boldsymbol{
ightarrow-\infty} \)

Problem 4

Equation: \( \boldsymbol{y=\frac{1}{2}(x - 1)^3-3} \)