/*! 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 65 Complete the square of each quad... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 65

Complete the square of each quadratic expression. Then graph each function using graphing techniques. $$ f(x)=x^{2}+2 x $$

Short Answer

Expert verified
Complete the square to get \( f(x) = (x + 1)^2 - 1 \) with vertex at \( (-1, -1) \). Graph opens upwards.

Step by step solution

01

- Identify constants

Given the quadratic function \( f(x) = x^2 + 2x \), identify the coefficients. Here, \( a = 1 \), \( b = 2 \).
02

- Find the term to complete the square

To complete the square, take the coefficient of \( x \) (which is 2), divide by 2, and square it: \( \left( \frac{2}{2} \right)^2 = 1 \).
03

- Form the perfect square trinomial

Add and subtract the square term inside the function: \( f(x) = x^2 + 2x + 1 - 1 \). This can be written as \( f(x) = (x + 1)^2 - 1 \).
04

- Rewrite the function

Rewrite the function in its vertex form: \( f(x) = (x + 1)^2 - 1 \). From this, the vertex is found to be \( (-1, -1) \).
05

- Graph the function

To graph \( f(x) \), plot the vertex \( (-1, -1) \) and note that the parabola opens upwards (since the leading coefficient is positive). Plot additional points if needed and draw the parabola.

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!

Key Concepts

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

Completing the Square
Completing the square is a method used to rewrite quadratic expressions in a form that makes them easier to graph and understand. It involves creating a perfect square trinomial from the quadratic expression. To complete the square for the function \( f(x) = x^2 + 2x \), follow these steps:

- Identify the coefficient of \( x \) (which is 2 in this case).
- Divide it by 2 and square the result: \( \left( \frac{2}{2} \right)^2 = 1 \).
- Add and subtract this square term from the function to form a perfect square: \( f(x) = x^2 + 2x + 1 - 1 \).

This creates a perfect square trinomial plus a constant term, which can then be factored and rewritten in a simpler form.
Vertex Form
The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \). This form provides an easy way to identify the vertex of the parabola, which is the point \((h, k)\).

For the function \( f(x) = (x + 1)^2 - 1 \):

- The expression inside the parenthesis, \( (x + 1) \), indicates a horizontal shift. Here, it's \( x + 1 = 0 \), so the vertex is shifted to the left by 1 unit (meaning \( h = -1 \)).
- The constant term \( -1 \) indicates a vertical shift down by 1 unit (so \( k = -1 \)).

Thus, the vertex of the parabola is at \( (-1, -1) \). The function in vertex form makes it straightforward to determine these shifts and plot the vertex.
Graphing Parabolas
Graphing a parabola involves plotting points and understanding the shape and direction of the curve. Starting with the vertex helps set a solid foundation for the graph.

For the function \( f(x) = (x + 1)^2 - 1 \):

- The vertex \( (-1, -1) \) is the minimum point of the parabola, as it opens upwards (since the leading coefficient is positive).
- To graph the parabola, plot the vertex first.
- Additional points can be found by selecting values for \( x \) and substituting them into the function to find corresponding \( y \)-values.

For example:
  • If \( x = 0 \), then \( f(0) = (0 + 1)^2 - 1 = 1 - 1 = 0 \). So, the point \( (0, 0) \) is on the graph.
  • If \( x = -2 \), then \( f(-2) = (-2 + 1)^2 - 1 = (-1)^2 - 1 = 0 \). So, the point \( (-2, 0) \) is on the graph.
Connect these points to form the parabola. Remember, the shape should be symmetrical about the vertex.

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

The equation \(y=(x-c)^{2}\) defines a family of parabolas, one parabola for each value of \(c .\) On one set of coordinate axes, graph the members of the family for \(c=0, c=3\), and \(c=-2\)

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=x^{4}-x^{2} \quad[-2,2]\)

Suppose \(f(x)=x^{3}+2 x^{2}-x+6\). From calculus, the Mean Value Theorem guarantees that there is at least one number in the open interval (-1,2) at which the value of the derivative of \(f\), given by \(f^{\prime}(x)=3 x^{2}+4 x-1\), is equal to the average rate of change of \(f\) on the interval. Find all such numbers \(x\) in the interval.

Rotational Inertia The rotational inertia of an object varies directly with the square of the perpendicular distance from the object to the axis of rotation. If the rotational inertia is \(0.4 \mathrm{~kg} \cdot \mathrm{m}^{2}\) when the perpendicular distance is \(0.6 \mathrm{~m},\) what is the rotational inertia of the object if the perpendicular distance is \(1.5 \mathrm{~m} ?\)

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=-x^{2}+3 x-2\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

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.