Problem 29 A function $f$ and a point \(P... [FREE SOLUTION]

91影视

Calculus Single Variable

Brian E. Blank, Steven G. Krantz

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 3: Problem 29

A function $f$ and a point $P$ are given. Find the slope-intercept form of the equation of the tangent line to the graph of $f$ at $P$. $$ f(x)=5 x^{2} \quad P=(2,20) $$

Short Answer

Expert verified

The equation is $ y = 20x - 20 $.

Step by step solution

Compute the Derivative of the Function

To find the slope of the tangent line, we need the derivative of the function. Given $ f(x) = 5x^2 $, we compute the derivative $ f'(x) $. The rule for derivatives of $ x^n $ is $ nx^{n-1} $, so here: $ f'(x) = 10x $.

Evaluate the Derivative at the Given Point

The slope of the tangent line at the point $ P = (2, 20) $ is the value of the derivative at $ x = 2 $. Substitute 2 into the derivative: $ f'(2) = 10 \times 2 = 20 $. So, the slope $ m $ of the tangent line is 20.

Use Point-Slope Form to Write the Tangent Line Equation

We use the point-slope form of the equation of a line: $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $(x_1, y_1)$ is the point $ P $. Substitute the values: $ y - 20 = 20(x - 2) $.

Convert to Slope-Intercept Form

To convert to slope-intercept form $ y = mx + b $, simplify and rearrange the equation: $ y - 20 = 20x - 40 $. Add 20 to both sides: $ y = 20x - 20 $. This is the equation of the tangent line in slope-intercept form.

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Key Concepts

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

Understanding Derivatives

In calculus, the derivative of a function is a measure of how the function values change as the input changes. It's key to understanding the concept of the slope of the tangent line. A tangent line is a straight line that touches a curve at one point without crossing it. To find a tangent line's slope, we use derivatives. For our function, $ f(x) = 5x^2 $, the derivative represents how $ f(x) $ changes as $ x $ changes.
The general rule for finding the derivative of a function $ x^n $ is $ nx^{n-1} $. For $ f(x) = 5x^2 $, we compute its derivative as $ f'(x) = 10x $.

The derivative, $ f'(x) = 10x $, gives us the slope at any point $ x $.
It's crucial because it tells us how steep the curve is at the given point.
By evaluating this at $ x = 2 $, we find that the slope of the tangent line at this point is 20.

Understanding the derivative is your first step in finding tangent lines!

Slope-Intercept Form of a Line

When writing the equation of a line, the slope-intercept form is a popular and straightforward way. It is expressed as $ y = mx + b $, where $ m $ represents the slope and $ b $ represents the y-intercept.
In our exercise, after calculating the slope with the derivative, you convert the tangent line equation into this form. Let's see why this form is so useful:

The slope $ m $ indicates how steep the line is, showing the rise over run.
The y-intercept $ b $ is where the line crosses the y-axis. It's a signature starting point for graphing lines.
Being able to rearrange any line into this form makes understanding its behavior intuitive and quick.

In the example, we have: $ y = 20x - 20 $. This shows a slope of 20 and a y-intercept of -20. With this, you can easily predict and graph the behavior of the tangent line.

Point-Slope Form

Sometimes the slope-intercept form is not directly applicable, especially when you know a point through which the line passes and its slope. Here comes the point-slope form, $ y - y_1 = m(x - x_1) $. It's perfect when you have:
- $ (x_1, y_1) $ as a known point on the line.
- $ m $ as the given slope of the line.
In our task, we had point $ P = (2, 20) $ and calculated slope $ m = 20 $.
This made our equation $ y - 20 = 20(x - 2) $, showcasing how the line relates to the point and its slope.
Using it helps:

Identify the relationship between the point and the slope directly.
Quickly form equations when dealing with specific coordinates rather than the y-intercept.
Give a clean and clear path from the derivative to the slope-intercept form.

Point-slope is your tool when the line's foundation (a point and slope) is known, making it a crucial step to adapt before moving to slope-intercept form.

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91影视

A function \(f\) and a point \(P\) are given. Find the slope-intercept form of the equation of the tangent line to the graph of \(f\) at \(P\). $$ f(x)=5 x^{2} \quad P=(2,20) $$

Short Answer

Step by step solution

Compute the Derivative of the Function

Evaluate the Derivative at the Given Point

Use Point-Slope Form to Write the Tangent Line Equation

Convert to Slope-Intercept Form

Key Concepts

Understanding Derivatives

Slope-Intercept Form of a Line

Point-Slope Form

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