Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to \(y=x^{2}-2 x+2\) at the point \((1,1)\).

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 1\).

Step 1: Identify the Function and Point

The given function is \(y = x^2 - 2x + 2\), and the point of interest is \((1, 1)\). We need the equation of a tangent line at this point.

Step 2: Find the Derivative

To find the slope of the tangent line, differentiate the function with respect to \(x\). \[\frac{dy}{dx} = \frac{d}{dx}(x^2 - 2x + 2) = 2x - 2\]

Step 3: Calculate the Slope of the Tangent Line

Evaluate the derivative at \(x = 1\) to find the slope of the tangent line:\[m = 2(1) - 2 = 0\]The slope at \((1, 1)\) is 0.

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of a line \(y - y_1 = m(x - x_1)\) with \((x_1, y_1) = (1, 1)\) and slope \(m = 0\):\[y - 1 = 0(x - 1)\]Simplifying this,\[y = 1\]

Key Concepts

Derivative Calculation

Calculating the derivative of a function is essential to determine the behavior of the function at any given point. To find the derivative of a function, we use the rules of differentiation, which provide a way to compute the rate at which the function's value is changing. In this problem, the function given is \(y = x^2 - 2x + 2\).
To find its derivative, we differentiate each term separately:

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(-2x\) is \(-2\).
• The derivative of a constant, 2, is 0.

When these are combined, the derivative of the function becomes \(\frac{dy}{dx} = 2x - 2\). The derivative function represents the slope of the tangent line at any point \(x\) on the graph of the original function.

Slope of Tangent Line

Once the derivative is calculated, finding the slope of the tangent line is straightforward. The derivative at a particular point gives us the exact slope of the function at that point. In this case, we need the slope where \(x = 1\).
By substituting \(x = 1\) into the derivative, \(2x - 2\), we get:

• \(m = 2(1) - 2 = 0\)

This tells us that at the point \( (1, 1) \), the slope of the tangent line is 0. A zero slope indicates that the tangent line is perfectly horizontal at that specific point on the curve.

Point-slope Form Equation

After determining the slope at a given point, we use the point-slope form to express the equation of the tangent line. The point-slope form is a simple way to write the equation of a line if you know one point on the line and the slope. It is expressed as:

• \(y - y_1 = m(x - x_1)\)

For our problem, the given point is \((1, 1)\) and the slope \(m = 0\). Substituting these into the point-slope form equation gives:

• \(y - 1 = 0(x - 1)\)

Solving the equation results in the simplified form \(y = 1\), which is the equation of the tangent line. This line is horizontal, as indicated by the slope of 0.

Differentiation

Differentiation is a fundamental concept in calculus that is used to compute the derivative of a function. It involves determining how a function changes as its input changes. This process allows us to find the slope of the function at any point, which is crucial for constructing tangent lines.
When you differentiate a function, you find the derivative, which is the formula for the function's slope at any point on its curve. In essence, differentiation provides a tool for analyzing and understanding the behavior of functions:

• Find instantaneous rates of change.
• Determine the slope of tangent lines to curves.
• Identify points of maximum or minimum on a graph.

Through differentiation, we gain insight into how a function behaves and interacts with its graph across different values of \(x\). In this exercise, differentiating \( y = x^2 - 2x + 2 \) allowed us to confirm that the slope of the tangent at the point \( (1, 1) \) was a neat 0, resulting in a simple, horizontal tangent line.