/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Graph the parabola \(f(x)=x^{2} ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 6

Graph the parabola \(f(x)=x^{2} .\) Explain why the secant lines between the points \((-a, f(-a))\) and \((a, f(a))\) have zero slope. What is the slope of the tangent line at \(x=0 ?\)

Short Answer

Expert verified
Answer: The secant lines between the points (-a, f(-a)) and (a, f(a)) have zero slope because both points have the same y-coordinate (a^2), which results in a horizontal line. The slope of the tangent line at x=0 for the parabola f(x) = x^2 is also 0, as the derivative f'(x) = 2x, and plugging in x=0 gives a slope of 0, indicating a horizontal tangent line at the vertex of the parabola.

Step by step solution

01

Graph the parabola f(x) = x^2

To graph the parabola \(f(x) = x^2\), we'll plot several points and draw the curve fitting these points. The points to be plotted are the values of x and their corresponding values of y (x^2). The graph will be a U-shaped curve, with the vertex (lowest point) at the origin (0,0).
02

Explain why the secant lines have zero slope

A secant line is a line that passes through two points on the graph. In this case, we need to explain why the secant lines passing through the points \((-a, f(-a))\) and \((a, f(a))\) have zero slopes. To do so, we need to calculate f(-a) and f(a). Given that \(f(x) = x^2\), we have: 1. \(f(-a) = (-a)^2 = a^2\) 2. \(f(a) = a^2\) Now, we have the points \((-a, a^2)\) and \((a, a^2)\). To find the slope of the secant line, we use the slope formula: slope = \(\frac{y_2 - y_1}{x_2 - x_1}\) In this case, we have: slope = \(\frac{a^2 - a^2}{a - (-a)}\) slope = \(\frac{0}{2a} = 0\) Since the slope of the secant line is 0, it means that the secant line is horizontal.
03

Determine the slope of the tangent line at x=0

To determine the slope of the tangent line at \(x=0\), we need to find the derivative of \(f(x) = x^2\). The derivative f'(x) represents the slope of the tangent line at any point x. The derivative of \(f(x) = x^2\) is: \(f'(x) = 2x\) Now, we have to find the slope of the tangent line at \(x = 0\). To do so, we plug 0 into the derivative function: \(f'(0) = 2(0) = 0\) The slope of the tangent line at \(x=0\) is 0, which means that the tangent line at the vertex of the parabola is horizontal.

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 91影视!

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 points at which the following functions \(f\) have discontinuities. For each point state the conditions in the continuity checklist that are violated.

Zero velocity A projectile is fired vertically upward and has a position given by \(s(t)=-16 t^{2}+128 t+192,\) for \(0 \leq t \leq 9\) a. Graph the position function, for \(0 \leq t \leq 9\) b. From the graph of the position function, identify the time at which the projectile has an instantaneous velocity of zero; call this time \(t=a\) c. Confirm your answer to part (b) by making a table of average velocities to approximate the instantaneous velocity at \(t=a\) d. For what values of \(t\) on the interval [0,9] is the instantaneous velocity positive (the projectile moves upward)? e. For what values of \(t\) on the interval [0,9] is the instantaneous velocity negative (the projectile moves downward)?

Sketch a possible graph of a function \(f\) that satisfies all of the given conditions. Be sure to identify all vertical and horizontal asymptotes. $$\lim _{x \rightarrow 0^{+}} f(x)=\infty, \lim _{x \rightarrow 0^{-}} f(x)=-\infty, \lim _{x \rightarrow \infty} f(x)=1$$, $$\lim _{x \rightarrow-\infty} f(x)=-2$$

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$

Suppose \(\lim _{x \rightarrow a} f(x)=L .\) Prove that \(\lim _{x \rightarrow a}[c f(x)]=c L,\) where \(c\) is a constant.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.