Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Finding an Equation of a Tangent Line In Exercises 81 and \(82,\) find an equation of the tangent line to the graph of the function \(f\) through the point \(\left(x_{0}, y_{0}\right)\) not on the graph. To find the point of tangency \((x, y)\) on the graph of \(f,\) solve the equation $$ f^{\prime}(x)=\frac{y_{0}-y}{x_{0}-x} $$ $$ \begin{array}{l}{f(x)=\frac{2}{x}} \\ {\left(x_{0}, y_{0}\right)=(5,0)}\end{array} $$

Short Answer

Expert verified
The equation of the tangent line to the graph of the function \(f(x) = \frac{2}{x}\) passing through the point \((5, 0)\) is \(y = -\frac{2x}{5} - 2.\)

Step by step solution

01

Find derivative of the function \(f(x)\)

The derivative of the function \(f(x) = \frac{2}{x}\) can be obtained using the power rule of derivative where \(f'(x) = nx^{n-1}\). The derivative of \(f(x)\) is thus \(f'(x) = -\frac{2} {x^{2}}.\)
02

Substitute in the provided equation

Now, we can use the equation \(f^{\prime}(x)=\frac{y_{0}-y}{x_{0}-x}\) substituting \((x_0, y_0) = (5, 0)\) and \(f'(x) = -\frac{2} {x^{2}}\). This gives us \(-\frac{2} {x^{2}} = \frac{-y}{5-x}.\)
03

Simplify to find the value of \(y\)

From the equation above, cross multiply to obtain the equation in form of \(y = mx + c\). By cross multiplication, we get \(y = 2 - \frac{2x}{5}.\)
04

Find y-intercept for the equation of tangent line

Set \(x = 5\) to find the y-intercept of the equation of the tangent line because this line passes through the point \((5, 0)\). Therefore, \(y = 2 - \frac{2(5)}{5} = -2 \). Thus, the equation of the tangent line is \(y = -\frac{2x}{5} -2.\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Derivative
In calculus, the derivative of a function essentially tells you how the function is changing at any given point. A derivative represents the slope of the tangent line to the curve of the function at a specific point. For example, when you differentiate a function, you are finding a new function that gives you the slope for any point on the original function's curve.

Understanding derivatives is crucial when solving problems involving rates of change, such as velocity and acceleration. It is also foundational in finding the tangent line to a curve, as we need the slope of the line at a particular point.
  • The derivative of a function is denoted as \(f'(x)\).
  • If \(f(x)\) is a function, \(f'(x)\) gives the slope of the tangent line at any point \(x\).
  • The process of finding the derivative is known as differentiation.
Thus, the derivative is a powerful tool in your mathematical toolbox, as it helps describe how a function behaves and changes at any point along its graph.
Power Rule
The power rule is a simple yet very powerful rule used to find the derivative of functions that are powers of \(x\). It's a shortcut that makes differentiation of polynomial functions straightforward and quick. The rule states:
  • If \(f(x) = x^n\), then the derivative \(f'(x)\) is \(nx^{n-1}\).
When applying the power rule, you multiply the original power by the coefficient of \(x\), and then subtract one from the original power.
  • Example: For \(f(x) = x^3\), the derivative using the power rule is \(f'(x) = 3x^2\).
  • This rule applies directly, simplifying calculations and allowing for quick results.
Through this rule, you efficiently develop the derivatives needed in solving for the tangent line equation.
Point of Tangency
The point of tangency is a key concept when dealing with tangent lines. It refers to the point where a tangent line touches the curve of a function without crossing it. This specific point is unique because the tangent line has the same slope as the function's curve at that exact location.

In problems where you need to find the equation of a tangent line, identifying the point of tangency is often a crucial step. The tangency point's coordinates are typically found by using the derivative to determine where the slope of the curve equals the slope of the line through a given external point.
  • The tangent line signifies the instantaneous rate of change at the point of tangency.
  • The derivative determines the slope of the tangent line at this specific point.
Thus, the point of tangency is essential for constructing the tangent line equation accurately.
Cross Multiplication
Cross multiplication is a method frequently used to solve equations involving fractions, especially when finding equations of lines like tangents. It allows you to eliminate the fractions by multiplying across the equal sign, helping in simplifying the equations.
  • Consider an equation like \(\frac{a}{b} = \frac{c}{d}\). By cross multiplying, you can reframe it to \(a \times d = b \times c\).
  • This technique is very useful in simplifying and solving rational equations.
When applying this in the context of finding a tangent line, cross multiplication helps isolate variables, making calculations more manageable.

As seen in the solution steps, when tackling \(-\frac{2}{x^{2}} = \frac{-y}{5-x}\), cross multiplication is employed to get the terms into a more workable form and ultimately find the equation for the tangent line.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=2 x^{2}+1 ; \frac{d x}{d t}=2 \text { centimeters per second }} \\ {\begin{array}{ll}{\text { (a) } x=-1} & {\text { (b) } x=0} & {\text { (c) } x=1}\end{array}}\end{array} $$

Conjecture Consider the functions \(f(x)=x^{2}\) and \(g(x)=x^{3} .\) (a) Graph \(f\) and \(f^{\prime}\) on the same set of axes. (b) Graph \(g\) and \(g^{\prime}\) 'on the same set of axes. (c) Identify a pattern between \(f\) and \(g\) and their respective derivatives. Use the pattern to make a conjecture about \(h^{\prime}(x)\) if \(h(x)=x^{n},\) where \(n\) is an integer and \(n \geq 2\) . (d) Find \(f^{\prime}(x)\) if \(f(x)=x^{4}\) . Compare the result with the conjecture in part (c). Is this a proof of your conjecture? Explain.

Finding a Second Derivative In Exercises \(91-98\) , find the second derivative of the function. $$ f(x)=\frac{x^{2}+3 x}{x-4} $$

True or False? In Exercises \(129-134\) , determine whether the statement is true or false. If is false, explain why or give an example that shows it is false. If the velocity of an object is constant, then its acceleration is zero.

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\frac{1}{1+x^{2}} ; \frac{d x}{d t}=6 \text { inches per second }} \\ {\begin{array}{ll}{\text { (a) } x=-2} & {\text { (b) } x=0} & {\text { (c) } x=2}\end{array}}\end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free