Problem 6 - The Calculus with Analytic Geometry

Question

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=x^{3}-x^{2}-x+10 $$

Step-by-Step Solution

Answer

Find the derivative, evaluate it at the given point, create a table with x, y, and slope (m), identify points with horizontal tangents by setting the derivative to zero, solve for x, find corresponding y-values, and sketch the graph.

Step 1: Find the Derivative

To find the slope of the tangent line to the graph, we first need to find the derivative of the function. Differentiate the function with respect to x:y = x^3 - x^2 - x + 10dy/dx = 3x^2 - 2x - 1

Step 2: Evaluate the Derivative at the Given Point

Given the point \(x_{1}, y_{1}\), substitute the x-coordinate into the derivative to find the slope (m) of the tangent line at that point:m = 3x_1^2 - 2x_1 - 1

Step 3: Create the Table of Values

Now, create a table of values for x, y, and m. Pick a variety of x-values and calculate the corresponding y-values using the original function, and the m-values using the derivative:| x | y | m ||---|---|---|| x_1 | y_1 | m || ... | ... | ... |Make sure to include points where the graph has a horizontal tangent.

Step 4: Identify Horizontal Tangents

Determine the points where the tangent is horizontal by setting the derivative equal to zero and solving for x:0 = 3x^2 - 2x - 1Solve this quadratic equation to find the x-coordinates of points with horizontal tangent.

Step 5: Solve for Horizontal Tangents

Factorize or use the quadratic formula \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ to find the x-values where the slope is horizontal. In this case:a = 3, b = -2, c = -1x = \[ \frac{2 \pm \sqrt{4 + 12}}{6} \] = \[ \frac{2 \pm \sqrt{16}}{6} \] = \[ \frac{2 \pm 4}{6} \]. List these x-values in the table.

Step 6: Find Corresponding y-values

For each x-value where the slope is horizontal, plug the x back into the original function to find the corresponding y-values:

Step 7: Draft the Table

Complete the table with x-values, their corresponding y-values, and the slopes (m):| x | y | m ||---|---|---|| x_1 | y_1 | m || ... | ... | ... |

Step 8: Sketch the Graph

Plot the function graph using a range of x-values. Draw the tangent lines at the given point and at the identified points with horizontal tangents to illustrate the slopes.

Key Concepts

derivative

To understand the slope of a tangent line to a graph, we need to first grasp the concept of a derivative. A derivative of a function at a particular point gives us the slope of the tangent line to the graph at that specific point. Mathematically, if we have a function \(y = f(x)\), the derivative, denoted as \(f'(x)\) or \(\frac{dy}{dx}\), is computed by differentiating the function with respect to \(x\). For instance, for the function given in the exercise, \(y = x^3 - x^2 - x + 10\), the process of differentiation yields the derivative as \(\frac{dy}{dx} = 3x^2 - 2x - 1\). This derivative represents how the value of y changes as x changes, giving us the slope of the tangent at any point \(x_{1}\) by evaluating \(3x_{1}^2 - 2x_{1} - 1\).

horizontal tangent

A horizontal tangent on the graph of a function occurs when the slope of the tangent line is zero. This means that the derivative of the function at that point is zero. To find such points, set the derivative equal to zero and solve for x. In our example, we set \(3x^2 - 2x - 1 = 0\). Solving this quadratic equation will provide the x-values where the slope is zero, indicating a horizontal tangent. Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), and \(c = -1\), we find \(x_1 = \frac{2 + 4}{6} = 1\) and \(x_2 = \frac{2 - 4}{6} = -\frac{1}{3}\). These points mark where the tangent to the curve is horizontal.

quadratic equation

Solving a quadratic equation is crucial in identifying points of horizontal tangents. A quadratic equation takes the form \(ax^2 + bx + c = 0\). The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), helps solve such equations. For the example in our exercise, substituting \(a = 3\), \(b = -2\), and \(c = -1\), we get \(x = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6}\), leading to two solutions: \(x = 1\) and \( x = -\frac{1}{3}\). Understanding the quadratic formula and solving these equations allows us to pinpoint where the derivative, and thus the slope of the tangent line, equals zero, indicating horizontal tangents on the graph.