Problem 26 - Calculus of a Single Variable

Question

Finding an Equation of a Tangent Line In Exercises \(25-32,(\text { a) find an equation of the tangent line to the graph of } f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ f(x)=x^{2}+2 x-1, \quad(1,2) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \( f(x)=x^{2}+2x-1 \) at the point (1,2) is \( y=4x-2 \).

Step 1: Determine the Derivative of the Function

The derivative of \( f(x) \) represents the slope of the tangent line at any given point on function. The derivative of \( f(x)=x^{2}+2x-1 \) can be calculated using power rule of differentiation i.e., if \( y=x^n \), then the derivative \( y'=nx^{n-1} \). The derivative is given by \( f'(x)=2x+2.\)

Step 2: Calculate the Slope of the Tangent Line

After finding the derivative of the function, substitute the x-value from the point of tangency (1,2) into the derivative to get the slope of the tangent line. Substituting \( x=1 \) into the derivative \( f'(x)=2x+2 \) will give you \( f'(1)=2*1+2=4 \). Therefore, the slope of the tangent line at (1,2) is 4.

Step 3: Determine the Equation of the Tangent Line

To find the equation of the tangent line, we use the point-slope form of linear equations, which is \( y-y_1=m(x-x_1) \). Here, \( m \) is the slope of the line, \( (x_1, y_1) \) is the point on the line. Given the slope \( m=4 \) and the point (1,2), the equation of the tangent line is \( y-2=4(x-1) \). Simplifying this equation gives \( y=4x-2 \).

Step 4: Graphing Function and Tangent Line

This part asks to graph the function and its tangent line at the point using graphing utility software, which is not possible here. But, the general idea would be to plot the function \( f(x) = x^{2} + 2x - 1 \) and then plot the tangent line \( y = 4x - 2 \) on the same graph.

Step 5: Confirm Results with Graphing Utility

This part asks to confirm the previously found results using a graphing utility with a derivative feature. Again, it is not possible to demonstrate this here. However, the key point is to ensure that the slope of the tangent line to the function at point (1,2) matches the calculated slope, which proves calculation was done correctly.

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It doesn't cross the curve, it merely skims its surface at the specified location.
When finding the equation of a tangent line, one crucial aspect is determining its slope at that particular point.

• The tangent line is important because it represents the instantaneous rate of change of the function at that point.
• In calculus, the tangent line is closely linked to the concept of a derivative, as the slope of this line is precisely the value of the derivative at that point.

For example, when given the function for a parabola, to find its tangent at a point, you use calculus techniques to determine its exact slope at the position of tangency.

Power Rule of Differentiation

The power rule is a fundamental tool in calculus, particularly when differentiating polynomial functions. It allows us to find the derivative of a term easily.
Simply put, the power rule states: if you have a function of the form \( y = x^n \), then its derivative is \( y' = nx^{n-1} \).

• This rule is incredibly handy because it simplifies the process of finding derivatives and therefore makes it straightforward to calculate slopes of tangent lines.
• For example, in the exercise, the function \( f(x) = x^2 + 2x - 1 \) uses the power rule. Each term's power is reduced by one, and it's multiplied by the original power, resulting in the derivative \( f'(x) = 2x + 2 \).

Understanding the power rule is essential for solving many calculus problems efficiently.

Derivative

The derivative of a function is a core concept in calculus. It measures how a function changes as its input changes, like its slope at any point.
Think of it as capturing the "rate of change" at a particular point.

• Calculating a derivative gives you a new function, which tells you the slope of the original function at any point along its curve.
• In our exercise, the derivative \( f'(x) = 2x + 2 \) tells us the slope of the function \( f(x) = x^2 + 2x - 1 \) at any value of \( x \).

By understanding and finding derivatives, you can determine how steep a curve is at any point, making it crucial for problems involving tangents and slopes.

Point-Slope Form

The point-slope form is a straightforward way to establish the equation of a line. It's very useful when you know a slope and a point on the line.
This form is given by: \( y - y_1 = m(x - x_1) \).

• In point-slope form, \( m \) is the slope, and \( (x_1, y_1) \) is a point on the line. Knowing these parameters makes it easy to generate the line's equation.
• For example, in the exercise, with slope \( m = 4 \) and point \( (1,2) \), we use this form to find the equation \( y - 2 = 4(x - 1) \), which simplifies to \( y = 4x - 2 \).

Applying point-slope form is a reliable method for quickly writing the equation of a line, given a specific point of tangency and its slope.