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Chapter 3: Problem 51

Find an equation of the line tangent to the graph of \(f\) at the given point. $$ f(x)=x^{2}-3 x-4 ;(2,-6) $$

Short Answer

Expert verified
The equation of the tangent line is \( y = x - 8 \).

Step by step solution

01

Find the derivative of the function

First, we need to find the derivative of the function \( f(x) = x^2 - 3x - 4 \). The derivative \( f'(x) \) can be found by differentiating each term with respect to \( x \). The derivative of \( x^2 \) is \( 2x \), the derivative of \( -3x \) is \( -3 \), and the derivative of \( -4 \) is \( 0 \). So, \( f'(x) = 2x - 3 \).
02

Evaluate the derivative at the given point

To find the slope of the tangent line at the point \((2,-6)\), plug \( x = 2 \) into the derivative \( f'(x) = 2x - 3 \). This gives \( f'(2) = 2(2) - 3 = 4 - 3 = 1 \). Thus, the slope of the tangent line is 1.
03

Use point-slope form of a line

Now that we have the slope of the tangent line, we can use the point-slope form to find the equation of the tangent line. The point-slope form is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the given point. Plugging in \( m = 1 \), \( x_1 = 2 \), and \( y_1 = -6 \): \( y - (-6) = 1(x - 2) \), which simplifies to \( y + 6 = x - 2 \).
04

Simplify the equation

The final step is to simplify the equation \( y + 6 = x - 2 \). Subtract 6 from both sides to get \( y = x - 8 \). This is the equation of the line tangent to the graph of the function at the point \((2, -6)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
In calculus, derivatives help us find the rate at which a function is changing at any given point. They are a critical tool for understanding tangent lines, as they provide the slope of such lines on the graph of a function. When given a polynomial function like \( f(x) = x^2 - 3x - 4 \), the process of differentiation requires applying rules to each term.
  • The power rule is used for terms like \( x^2 \), which states that the derivative of \( x^n \) is \( nx^{n-1} \).
  • The derivative of constants like \( -4 \) is zero since constants don't change.
  • For linear terms like \( -3x \), the derivative is simply the coefficient \( -3 \).
Understanding the derivative, \( f'(x) = 2x - 3 \), allows you to compute the slope of the tangent line at any point \( x \) on the curve. By plugging the given \( x \)-coordinate into this derivative, you determine how steep the tangent line is at that particular location.
Slope of a Line
The slope of a line is an indicator of how steep the line is, and in calculus, this translates to how steep a tangent line is at a given point on a graph. For a linear function, the slope is constant, but for non-linear functions, it can vary.
  • Here, the slope of the tangent at \((2,-6)\) is found by evaluating the derivative at \(x = 2\).
  • This evaluation gives \( f'(2) = 2 \times 2 - 3 = 1 \), meaning the slope of the tangent line at \(x = 2\) is \(1\).
The slope is crucial because it determines the direction and steepness of the tangent line. A positive slope like \(1\) means the line ascends from left to right, providing a great way to visualize how the function is behaving at the point of tangency.
Point-Slope Form
The point-slope form is an equation that allows us to construct the equation of a line when we know the slope and a point on the line. It is very useful in calculus, especially when finding tangent lines.
  • The point-slope form is given by \( y - y_1 = m(x - x_1) \), where \( m \) stands for the slope and \((x_1, y_1)\) represents a specific point on the line.
  • In our exercise, \( m = 1 \), \( x_1 = 2 \), and \( y_1 = -6 \).
Substituting these values into the point-slope formula gives \( y + 6 = 1(x - 2) \). Simplifying this to standard form, \( y = x - 8 \), reveals the line equation in a more straightforward way. This method effectively bridges the concepts of derivatives and line equations, providing a tangible result.

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Most popular questions from this chapter

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