Problem 40 - Brief Applied Calculus

Question

a. Find the equation of the tangent line to \(f(x)=x^{2}-4 x+6\) at \(x=1\) b. Graph the function and the tangent line on the window \([-1,5]\) by \([-2,10]\).

Step-by-Step Solution

Answer

The tangent line at \( x = 1 \) is \( y = -2x + 5 \) and can be graphed alongside the function \( f(x) = x^2 - 4x + 6 \) in the specified window.

Step 1: Differentiate the Function

To find the equation of the tangent line, we first need to find the derivative of the function, which will give us the slope of the tangent line. The given function is \( f(x) = x^2 - 4x + 6 \). Differentiating with respect to \( x \), we get: \( f'(x) = 2x - 4 \).

Step 2: Evaluate the Derivative at the Point

Evaluate the derivative \( f'(x) = 2x - 4 \) at \( x = 1 \) to find the slope of the tangent line. Substitute \( x = 1 \) into the derivative: \( f'(1) = 2(1) - 4 = 2 - 4 = -2 \). The slope of the tangent line is \( -2 \).

Step 3: Find the Function Value at the Point

Substitute \( x = 1 \) into the original function to find the value of the function at this point. This will give us a point on the tangent line. \( f(1) = 1^2 - 4(1) + 6 = 1 - 4 + 6 = 3 \). So the point is \( (1, 3) \).

Step 4: Write the Equation of the Tangent Line

Using the point-slope form of the equation for a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point, substituting \( m = -2 \) and \( (1, 3) \), we have: \( y - 3 = -2(x - 1) \). Simplifying gives: \( y = -2x + 5 \). Thus, the equation of the tangent line is \( y = -2x + 5 \).

Step 5: Plot the Function and Tangent Line

To graph the function \( f(x) = x^2 - 4x + 6 \) and the tangent line \( y = -2x + 5 \) in the window \([-1, 5]\) by \([-2, 10]\), plot the curve of the quadratic function and the line using the calculated equations, making sure to show the point where the tangent touches the curve at \( x = 1 \). Ensure both the function and tangent line fit within the given window range to visualize their relationship.

Key Concepts

Derivative

When dealing with functions, a derivative allows you to determine the rate of change at any given point. Specifically, it shows how a small change in the input (\( x \)) affects the output (\( f(x) \)).To find the derivative, we apply differentiation rules to the function's formula. For example, if we have \( f(x) = x^2 - 4x + 6 \), the derivative, found through the power rule, is \( f'(x) = 2x - 4 \).
In essence, \( f'(x) \) gives the slope of the tangent to the curve at any point \( x \). This is crucial because it connects the derivative directly to the tangent line, as the slope tells us how steep or flat the line is at that specific point on the curve.

• Power Rule: Differentiate each term independently.
• Linear Functions: The slope of simple lines.

Differentiating helps us understand the dynamics of the function, predicting behavior and inferring key features like critical points.

Slope of Tangent Line

The slope of the tangent line is a fundamental concept in calculus, represented by the derivative value at a specific point. It's essentially how slanted the line that "just touches" the curve is.In practice, once the derivative is known (\( f'(x) = 2x - 4 \) for our given function), we substitute the specific \( x \) value where the slope is needed. For instance, if we want the slope at \( x = 1 \), we substitute it into the derivative equation:
\[f'(1) = 2(1) - 4 = -2\]This calculation tells us that the slope of the tangent line at \( x = 1 \) is \( -2 \).

• Negative Slope: Line falls as you move to the right.
• Zero Slope: Horizontal line, touches the curve without crossing it.
• Positive Slope: Line rises as you move to the right.

Understanding the slope helps in determining the direction and steepness of the tangent line, making it invaluable for problems involving rates and changes.

Point-Slope Form

Point-slope form is a method used to write the equation of a line given its slope and a point it passes through. This approach simplifies the process of finding the tangent line equation.The formula is given by:
\[y - y_1 = m(x - x_1)\]Where:

• \( m \) is the slope.
• \( (x_1, y_1) \) is a point on the line.

For the tangent line at \( x = 1 \) of \( y = -2x + 5 \), we have:

• Slope (\( m \)) at point \( x = 1 \): \( -2 \)
• Point on the curve: \( (1, 3) \)

Substitute these into the formula:
\[y - 3 = -2(x - 1)\]Solving, we get:
\[y = -2x + 5\]This gives us the equation of the tangent line in slope-intercept form, indicating its linear nature along with where it intersects the axes.

Graphing Functions

Graphing functions is a powerful visual method to understand both the original curve, like \( f(x) = x^2 - 4x + 6 \), and its associated tangent at a particular point.Visualizing these functions helps to comprehend their intersection and behavior. To graph, we adopt the following steps:

• Plot the quadratic function over a determined range, such as \([-1, 5]\).
• Mark the tangent line \( y = -2x + 5 \), ensuring it passes smoothly through \( (1, 3) \).

These visuals should fit within a chosen window, perhaps \([-2, 10]\) on the \( y \)-axis and show the function's parabolic shape with the linear tangent line intersecting at \( x = 1 \).Through graphing, not only do we get a better grasp of theoretical results, but we also perceive how different mathematical elements relate visually. It is crucial for verifying the correctness of calculations and understanding dynamic changes.