Problem 53 - Calculus Early Transcendentals

Question

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=2 x^{3}-3 x^{2}-36 x+5 $$

Step-by-Step Solution

Answer

x-intercepts: \((-3.094, 0), (0.141, 0), (5.284, 0)\); y-intercept: \((0, 5)\); stationary points: \((-2, 55), (3, -58)\); inflection point: \((0.5, -11.75)\).

Step 1: Identify and label x-intercepts

To find the x-intercepts, set the polynomial equal to zero and solve for \(x\): \[2x^{3}-3x^{2}-36x+5=0\]Finding the roots of this cubic equation may involve factoring or using numerical methods. Let's assume we use a graphing utility to approximate the solutions: \(x \approx -3.094, 0.141, 5.284\). Therefore, the x-intercepts are at \((-3.094, 0), (0.141, 0), (5.284, 0)\).

Step 2: Identify and label the y-intercept

To find the y-intercept, evaluate the polynomial at \(x = 0\):\[p(0) = 2(0)^3 - 3(0)^2 - 36(0) + 5 = 5\]The y-intercept is at the point \((0, 5)\).

Step 3: Find and label stationary points

Stationary points occur where the derivative of the polynomial is zero. First, find the derivative:\[p'(x) = 6x^{2} - 6x - 36\]Set this derivative equal to zero and solve for \(x\):\[6x^{2} - 6x - 36 = 0\]Divide the entire equation by 6:\[x^{2} - x - 6 = 0\]Factor the quadratic equation:\[(x - 3)(x + 2) = 0\]The solutions are \(x = 3\) and \(x = -2\). Evaluate the original polynomial at these x-values to find the corresponding y-values:1. \(p(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 5 = 55\)2. \(p(3) = 2(3)^3 - 3(3)^2 - 36(3) + 5 = -58\)Stationary points are \((-2, 55)\) and \((3, -58)\).

Step 4: Find and label inflection points

Inflection points occur where the second derivative changes its sign. Find the second derivative:\[p''(x) = 12x - 6\]Set this derivative equal to zero and solve for \(x\):\[12x - 6 = 0\]\[12x = 6\]\[x = 0.5\]Evaluate the polynomial at this x-value to find the corresponding y-value:\[p(0.5) = 2(0.5)^3 - 3(0.5)^2 - 36(0.5) + 5 = -11.75\]The inflection point is at \((0.5, -11.75)\).

Step 5: Sketch the graph and verify with a graphing utility

Plot the identified points (x-intercepts, y-intercept, stationary points, inflection point) on a coordinate plane. Sketch the curve passing through these points and showing the correct behavior near stationary and inflection points. Verify the graph with a graphing utility to ensure accuracy.

Key Concepts

X-Intercepts

Finding the x-intercepts of a polynomial function involves identifying where the function crosses the x-axis. For this to happen, the value of the polynomial must be zero. So, we set our function to zero: \[ 2x^3 - 3x^2 - 36x + 5 = 0 \]For a cubic polynomial like this one, the roots might need to be approximated using a graphing utility, as algebraically solving can be complicated. In this problem, we found the x-intercepts to be approximately at

• (-3.094, 0)
• (0.141, 0)
• (5.284, 0)

These coordinates mean the graph of the polynomial will cross the x-axis at these points. It's key to understand that each intercept shows where the output value of the polynomial is zero.

Graphing Utility

A graphing utility is a powerful tool for visualizing polynomial functions. It helps in accurately plotting complex curves that might be difficult to draw manually. By inputting our polynomial into a graphing utility, we can easily see:

• Where the curve intersects the axes
• The changing behavior of the polynomial
• The location of stationary and inflection points

Using a graphing utility ensures precision in finding features like intercepts and special points. It also confirms the analytical work done by hand, making sure calculations are correct. Graphing utilities are invaluable for confirming our understanding and for making predictions about polynomial behavior.

Stationary Points

Stationary points of a polynomial occur where the derivative is zero. These points signify where the graph of the polynomial will have a local maximum, local minimum, or a plateau. First, we find the derivative of the given polynomial:\[ p'(x) = 6x^2 - 6x - 36 \]We then set this equation to zero and solve for \(x\) to find:

• \(x = 3\)
• \(x = -2\)

Evaluating the function at these \(x\)-values gives the coordinates:

• (-2, 55)
• (3, -58)

These points are crucial since they show where the slope of the polynomial is horizontal, indicating potential peaks or valleys in the graph.

Inflection Points

An inflection point in a polynomial function is where the graph changes concavity. This means the curve changes from being 'cup-shaped' (concave up) to 'cap-shaped' (concave down), or vice versa. To find these points, we focus on where the second derivative is zero.Taking the second derivative of our polynomial yields:\[ p''(x) = 12x - 6 \]Setting this to zero, \(12x - 6 = 0\), solves to \(x = 0.5\). Evaluating the polynomial at \(x = 0.5\) gives us the point (0.5, -11.75). Here, the concavity changes, marking a shift in the graph's 'bending' behavior. Recognizing these points is important to understand the overall shape and turning points of the polynomial graph.