/*! 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 30 Write the quadratic function in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 30

Write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$f(x)=-x^{2}-4 x+1$$

Short Answer

Expert verified
The standard form of the quadratic function is \(f(x) = -(x+2)^2 + 5\). The vertex is at \((-2, 5)\), the axis of symmetry is \(x = -2\), and the x-intercepts are \(-3\) and \(1\). The graph of this function is a downward-opening parabola.

Step by step solution

01

Transform to standard form

Complete the square on the function to transform it to standard form. Rearrange the function to get \(f(x)= -(x^2 + 4x) + 1\). The standard form of a quadratic function is \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. Complete the square inside the parenthesis: \(f(x)= -[(x^2 + 4x + 4) - 4] + 1 = -(x+2)^2 + 5\). So, the standard form is \(f(x) = -(x+2)^2 + 5\).
02

Identify the vertex and axis of symmetry

From the standard form \(f(x) = -(x+2)^2 + 5\), the vertex of the function is \((-2, 5)\). Since the equation of the axis of symmetry of a parabola \(f(x) = a(x-h)^2 + k\) is \(x = h\), the axis of symmetry of this function is \(x = -2\).
03

Identify the x-intercepts

To find the x-intercepts, set the function to zero and solve for \(x\). This gives us \(0 = -(x+2)^2 + 5\). Solving for \(x\) results in \(x = -1 \pm \sqrt{4}\) which simplifies to \(x = -1 \pm 2\). Therefore, the x-intercepts are \(-3\) and \(1\).
04

Sketch the graph

Begin by plotting the vertex at point \((-2, 5)\) and draw the axis of symmetry at \(x = -2\). Mark the x-intercepts at \(-3\) and \(1\) on the x-axis. The graph opens downward because the value of \(a\) in the standard form \(f(x) = -(x+2)^2 + 5\) is negative. Draw a smooth curve to connect these points and extend the arms of the parabola down on both sides because the function is not restricted in the domain.

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

A rectangular parking lot with a perimeter of 440 feet is to have an area of at least 8000 square feet. Within what bounds must the length of the rectangle lie?

Match the cubic function with the numbers of rational and irrational zeros. (a) Rational zeros: \(0 ;\) irrational zeros: 1 (b) Rational zeros: \(3 ;\) irrational zeros: 0 (c) Rational zeros: \(1 ;\) irrational zeros: 2 (d) Rational zeros: \(1 ;\) irrational zeros: 0 $$f(x)=x^{3}-x$$

Write the polynomial as the product of linear factors and list all the zeros of the function. $$g(x)=x^{4}-4 x^{3}+8 x^{2}-16 x+16$$

Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{3}{2} x^{2}-\frac{23}{2} x+6=\frac{1}{2}\left(2 x^{3}-3 x^{2}-23 x+12\right)$$

Determine whether the statement is true or false. Justify your answer. $$i^{44}+i^{150}-i^{74}-i^{109}+i^{61}=-1$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

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.