Graphing Cubic Functions

Problem: Let \( f \) be a cubic function given by \( f(x) = x^3 \).

1. Find the \( x \)- and \( y \)-intercepts of the graph of \( f \).
2. Find the domain and range of \( f \).
3. Sketch the graph of \( f \).

Solution:

1. The \( y \)-intercept is given by \( (0, f(0)) = (0, 0) \).
   The \( x \)-coordinates of the \( x \)-intercepts are solutions to \( x^3 = 0 \), which gives the point \( (0, 0) \).
2. The domain of \( f \) is the set of all real numbers \( (-\infty, +\infty) \). Since the leading coefficient of \( x^3 \) is positive (\( a = 1 \)), the graph rises on the right and falls on the left, making the range the set of all real numbers \( (-\infty, +\infty) \).
3. Construct a table of values and plot:

   \( x \)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
   \( f(x) = x^3 \)	\(-8\)	\(-1\)	\(0\)	\(1\)	\(8\)

   Note: Since \( f(-x) = (-x)^3 = -x^3 = -f(x) \), function \( f \) is odd, and its graph is symmetric with respect to the origin \( (0, 0) \).

   

Problem: Let \( f \) be a cubic function given by \( f(x) = -(x - 2)^3 \).

1. Find the \( y \)-intercept of the graph of \( f \).
2. Find all zeros of \( f \) and their multiplicity.
3. Find the domain and range of \( f \).
4. Use the intercepts and other properties to sketch the graph of \( f \).

Solution:

1. The \( y \)-intercept is given by \( (0, f(0)) = (0, -(0 - 2)^3) = (0, -(-8)) = (0, 8) \).
2. The zeros of \( f \) are solutions to \( -(x - 2)^3 = 0 \). Function \( f \) has one zero at \( x = 2 \) of multiplicity \( 3 \), meaning the graph crosses the \( x \)-axis at \( x = 2 \) with a flattening inflection point.
3. Expanding \( f(x) \): \[ f(x) = -(x - 2)^3 = -x^3 + 6x^2 - 12x + 8 \] The leading coefficient of \( x^3 \) is negative (\( a = -1 \)), so the graph falls on the right and rises on the left. The range is the set of all real numbers \( (-\infty, +\infty) \).
4. The graph cuts the \( x \)-axis at \( x = 2 \) and passes through the \( y \)-intercept \( (0, 8) \). Geometrically, \( f(x) = -(x - 2)^3 \) is the graph of \( y = x^3 \) shifted \( 2 \) units to the right and reflected across the \( x \)-axis.

   

Problem: Let \( f \) be a cubic function given by \( f(x) = x^3 + 2x^2 - x - 2 \).

1. Factor \( f(x) \).
2. Find all zeros of \( f \) and their multiplicity.
3. Find the domain and range of \( f \).
4. Use intercepts and properties to sketch the graph.

Solution:

1. Factor by grouping: \[ f(x) = x^2(x + 2) - 1(x + 2) = (x + 2)(x^2 - 1) = (x + 2)(x - 1)(x + 1) \]
2. The zeros of \( f \) are solutions to \( (x + 2)(x - 1)(x + 1) = 0 \), yielding \( x = -2 \), \( x = 1 \), and \( x = -1 \), each with multiplicity \( 1 \).
3. The domain of \( f \) is \( (-\infty, +\infty) \). The leading coefficient is positive, so the range is \( (-\infty, +\infty) \).
4. The \( y \)-intercept is \( (0, f(0)) = (0, -2) \). The graph cuts the \( x \)-axis at \( x = -2, -1, \) and \( 1 \).

   

Problem: Let \( f \) be given by \( f(x) = -x^3 + 3x + 2 \).

1. Show that \( (x - 2) \) is a factor of \( f(x) \) and factor \( f(x) \) completely.
2. Find all zeros of \( f \) and their multiplicity.
3. Find the domain and range of \( f \).
4. Use intercepts and properties to sketch the graph.

Solution:

1. Dividing \( f(x) \) by \( x - 2 \) yields quotient \(-x^2 - 2x - 1\) and remainder \(0\). Thus: \[ f(x) = (x - 2)(-x^2 - 2x - 1) = -(x - 2)(x^2 + 2x + 1) = -(x - 2)(x + 1)^2 \]
2. Setting \( -(x - 2)(x + 1)^2 = 0 \), we find zeros at \( x = 2 \) (multiplicity \( 1 \), crossing axis) and \( x = -1 \) (multiplicity \( 2 \), tangent to axis).
3. Domain is \( (-\infty, +\infty) \). Leading coefficient is negative, so range is \( (-\infty, +\infty) \).
4. The \( y \)-intercept is \( (0, 2) \). The graph cuts the \( x \)-axis at \( x = 2 \) and is tangent to it at \( x = -1 \).