Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 5

Calculate the equation of the tangents to \(y=9-x^{2}\) at the points where \(y\) crosses the \(x\) axis.

Short Answer

Expert verified
Answer: The equations of the tangents are \(y = -6x + 18\) and \(y = 6x + 18\).

Step by step solution

01

Find the points of intersection with the x-axis

To find the points where the curve intersects the x-axis, we set \(y\) to 0 and solve for \(x\): $$ 0 = 9 - x^2 $$ Now, we should find the values of \(x\) that satisfy the above equation. $$ x^2 = 9 $$ So, \(x = \pm 3\) are the x-intercepts of the curve. And the corresponding points are \((3,0)\) and \((-3,0)\).
02

Find the derivatives of the given function

Now, we will find the derivative of \(y\) with respect to \(x\) to determine the slope of the tangent at these points. $$ y = 9 - x^2 $$ Differentiating both sides with respect to \(x\), we get $$ \frac{dy}{dx} = -2x $$
03

Evaluate the slopes at the points of intersection

Now, we will find the slopes of the tangents at the points \((3,0)\) and \((-3,0)\) by plugging in the x-coordinates into the derivative we found earlier. For the point \((3,0)\): $$ \frac{dy}{dx} = -2(3) = -6 $$ For the point \((-3,0)\): $$ \frac{dy}{dx} = -2(-3) = 6 $$
04

Write the equations for the tangents using the point-slope form

Now that we have the slopes at the points of intersection, we can use the point-slope form of a line to write the equations for the tangents. The point-slope form of a line is given by: $$ y - y_1 = m(x - x_1) $$ Where \((x_1, y_1)\) is the point on the line, and \(m\)is the slope. For the tangent at point \((3,0)\) with slope \(-6\): $$ y - 0 = -6(x - 3) \\ y = -6x + 18 $$ For the tangent at point \((-3,0)\) with slope \(6\): $$ y - 0 = 6(x + 3) \\ y = 6x + 18 $$ So the equations of the tangents to the curve \(y = 9 - x^2\) at the points where it intersects the x-axis are \(y = -6x + 18\) and \(y = 6x + 18\).

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.

Start your free trial

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

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

Determine the location of all maximum and minimum points of the following functions: (a) \(y=x \mathrm{e}^{x}\) (b) \(y=\frac{1}{1+x^{2}}\) (c) \(y=\frac{x^{5}}{5}-\frac{x^{3}}{3}\) (d) \(y=\frac{x^{4}}{4}+x+1\) (e) \(y=(1-\ln x) x\)

Find the rate of change of $$ q(t)=2 \mathrm{e}^{-t / 2} \cos 2 t $$ when \(t=1\)

Find the equation of the tangent to \(y=x \mathrm{e}^{x}\) where \(x=1\).

Use logarithmic differentiation to differentiate $$ y=x^{x} $$

Calculate \(y^{\prime}\) where \(y\) is given by (a) \(\left(t^{2}+1\right) \sin 4 t\) (b) \((3 t+7) \mathrm{e}^{-2 t}\) (c) \(\left(e^{x}+\mathrm{e}^{-2 \mathrm{n}}\right)\left(3 x^{2}-2 x\right)\) (d) \(\sqrt{x} e^{x}\) (e) \(\frac{t^{2}+1}{\mathrm{e}^{t}}\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

Read Explanation

Physical Quantities And Units

Read Explanation

Physics of Motion

Read Explanation

Linear Momentum

Read Explanation
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