/*! 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

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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.

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Most popular questions from this chapter

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\sqrt{|x|}-\sqrt{|x-1|}$$

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

a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval. b. Use a graphing utility to find all the solutions to the equation on the given interval. c. Illustrate your answers with an appropriate graph. $$x^{3}-5 x^{2}+2 x=-1 ;(-1,5)$$

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)?

Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. $$f(x)=\frac{1}{\sqrt{x} \sec x}$$
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