Chapter 1: Problem 21
Find an equation of the tangent line to the graph of \(f(x)=x^{2}-2 x\) at (a) (-2,8) (b) (1,-1) (c) (4,8) .
Short Answer
Expert verified
(a) y = -6x - 4, (b) y = -1, (c) y = 6x - 16.
Step by step solution
01
Differentiate the Function
The first step is to find the derivative of the function \( f(x) = x^2 - 2x \). The derivative, \( f'(x) \), will give us the slope of the tangent line at any point \( x \) on the curve. Using basic differentiation rules, we have:\[f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x) = 2x - 2.\]
02
Calculate Slope at Point (a): (-2,8)
Substitute \( x = -2 \) into \( f'(x) \) to find the slope of the tangent line at the point \((-2, 8)\):\[f'(-2) = 2(-2) - 2 = -4 - 2 = -6.\]
03
Equation of Tangent Line at Point (a)
Use the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1) = (-2, 8)\).The equation becomes:\[y - 8 = -6(x + 2).\]Simplifying, we get:\[y = -6x - 12 + 8 \Rightarrow y = -6x - 4.\]
04
Calculate Slope at Point (b): (1,-1)
Substitute \( x = 1 \) into \( f'(x) \):\[f'(1) = 2(1) - 2 = 2 - 2 = 0.\]
05
Equation of Tangent Line at Point (b)
The slope \( m \) is 0, indicating a horizontal tangent line.Using the point \((1, -1)\):\[y - (-1) = 0(x - 1) \Rightarrow y + 1 = 0 \Rightarrow y = -1.\]
06
Calculate Slope at Point (c): (4,8)
Substitute \( x = 4 \) into \( f'(x) \):\[f'(4) = 2(4) - 2 = 8 - 2 = 6.\]
07
Equation of Tangent Line at Point (c)
Use the point-slope form with the point \((4, 8)\):\[y - 8 = 6(x - 4).\]Simplifying, we get:\[y = 6x - 24 + 8 \Rightarrow y = 6x - 16.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
A derivative is a fundamental tool in calculus that measures how a function changes as its input changes. You can think of it as a way to find the rate at which things are happening, such as speed or growth.
In our exercise, the function given is \( f(x) = x^2 - 2x \). To find the derivative, we apply differentiation rules to understand how fast \( f(x) \) is changing at any value of \( x \).
Interestingly, when we differentiate \( f(x) = x^2 - 2x \), we get \( f'(x) = 2x - 2 \). This new function, \( f'(x) \), acts as a machine that spits out the slope of the tangent line at any point you choose.
Imagine driving on a roller coaster; the derivative tells you how steep or flat the track is at any point!
In our exercise, the function given is \( f(x) = x^2 - 2x \). To find the derivative, we apply differentiation rules to understand how fast \( f(x) \) is changing at any value of \( x \).
Interestingly, when we differentiate \( f(x) = x^2 - 2x \), we get \( f'(x) = 2x - 2 \). This new function, \( f'(x) \), acts as a machine that spits out the slope of the tangent line at any point you choose.
Imagine driving on a roller coaster; the derivative tells you how steep or flat the track is at any point!
Slope of a Tangent
The slope of a tangent line is an expression of how slanted a line is relative to the horizontal. It tells you whether the line is rising, falling, or staying flat.
In the world of mathematics, when we speak of the slope of a tangent, we're indulging in the notion that the tangent just brushes a single point on the curve, sending us a secret message about the curve’s behavior right there!
To find it, we use the derivative: for each point on our function \( f(x) = x^2 - 2x \), we can substitute the \( x \)-value into \( f'(x) = 2x - 2 \) to unveil the slope.
In the world of mathematics, when we speak of the slope of a tangent, we're indulging in the notion that the tangent just brushes a single point on the curve, sending us a secret message about the curve’s behavior right there!
To find it, we use the derivative: for each point on our function \( f(x) = x^2 - 2x \), we can substitute the \( x \)-value into \( f'(x) = 2x - 2 \) to unveil the slope.
- For instance, at \((-2,8)\), the slope is \(-6\).
- At \((1,-1)\), it's \(0\), meaning the tangent is perfectly horizontal.
- While at \((4,8)\), it becomes \(6\), showing a steep rise.
Point-Slope Form
The point-slope form is a clever tool that allows you to write the equation of a line if you know a point on the line and its slope. The formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is a point on the line.
This form is extremely useful because it gets us directly to the line's equation without much hustle. You start from a known point and build the line around it.
Let's see how it shines:
This form is extremely useful because it gets us directly to the line's equation without much hustle. You start from a known point and build the line around it.
Let's see how it shines:
- For \((-2,8)\), with a slope of \(-6\), we write \( y - 8 = -6(x + 2) \).
- For \((1,-1)\), a horizontal slope of \(0\) leads to \( y = -1 \).
- Then, for \((4,8)\), a slope of \(6\) gives us \( y - 8 = 6(x - 4) \).
Differentiation Rules
Differentiation rules are like a set of handy instructions in a calculus toolkit. They help to find derivatives quickly and accurately, guiding us to understand the rate of change problems without getting lost in the math.
Fundamental rules include:
Fundamental rules include:
- **Power Rule**: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- **Constant Rule**: The derivative of a constant is \(0\).
- **Sum/Difference Rule**: The derivative of two functions added or subtracted is the sum or difference of their derivatives.