/*! 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 11 A quadratic function is given. (... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 11

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x\) - and \(y\) -intercept(s). (c) Sketch its graph. $$f(x)=2 x^{2}+6 x$$

Short Answer

Expert verified
The quadratic function is already in standard form. The vertex is \((-\frac{3}{2}, -\frac{9}{2})\), with \(x\)-intercepts \((0, 0)\), \((-3, 0)\), and a \(y\)-intercept \((0, 0)\).

Step by step solution

01

Express in Standard Form

The standard form of a quadratic function is \( ax^2 + bx + c \). The given function \( f(x) = 2x^2 + 6x \) already looks like it's in standard form where \( a = 2 \), \( b = 6 \), and \( c = 0 \).
02

Find the Vertex

The vertex of a quadratic function in standard form \( ax^2 + bx + c \) is found using \( x = \frac{-b}{2a} \). For \( f(x) = 2x^2 + 6x \), \( a = 2 \) and \( b = 6 \). Therefore, \( x = \frac{-6}{2 \times 2} = -\frac{3}{2} \). Substitute \( x = -\frac{3}{2} \) back into the function to find \( y \): \( y = 2(-\frac{3}{2})^2 + 6(-\frac{3}{2}) = 2\times\frac{9}{4} - 9 = \frac{9}{2} - 9 = -\frac{9}{2} \). Thus, the vertex is \((-\frac{3}{2}, -\frac{9}{2})\).
03

Find the x-intercepts

To find the \( x \)-intercepts, set \( f(x) = 0 \). So, \( 2x^2 + 6x = 0 \). Factor the equation: \( 2x(x + 3) = 0 \). Set each factor to zero: \( 2x = 0 \) or \( x + 3 = 0 \). Solving these gives \( x = 0 \) or \( x = -3 \). Therefore, the \( x \)-intercepts are \( (0, 0) \) and \( (-3, 0) \).
04

Find the y-intercept

The \( y \)-intercept is found by setting \( x = 0 \) in the function. Substituting gives \( y = 2(0)^2 + 6(0) = 0 \). Thus, the \( y \)-intercept is \( (0, 0) \).
05

Sketch the Graph

The function \( f(x) = 2x^2 + 6x \) is a parabola opening upwards because \( a = 2 \) is positive.Using the vertex \((-\frac{3}{2}, -\frac{9}{2})\), the \( x \)-intercepts \((0, 0)\) and \((-3, 0)\), and the \( y \)-intercept \((0, 0)\), sketch a symmetric parabola around the vertex, indicating these points on the graph.

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.

Standard Form of a Quadratic
A quadratic function can beautifully illustrate the shape of a parabola on a graph. It often appears in the standard form, expressed as \( ax^2 + bx + c \), where:
  • \( a \) is the coefficient that determines the degree of the parabola鈥檚 width and direction.
  • \( b \) influences the symmetry and linear component.
  • \( c \) is the constant term that moves the parabola up or down on the graph.
For the function \( f(x) = 2x^2 + 6x \), the standard form is clear. Here, \( a = 2 \), \( b = 6 \), and \( c = 0 \). This standard appearance tells us important information about the function and prepares us to find other critical points like the vertex and intercepts.
Vertex of a Parabola
The vertex of a parabola is a key point. It is the highest or lowest point of the parabola depending on the direction it opens.To find the vertex for a quadratic in standard form, you use the formula \( x = \frac{-b}{2a} \). For our function \( f(x) = 2x^2 + 6x \), with \( a = 2 \) and \( b = 6 \), plug in the values:\[x = \frac{-6}{2 \times 2} = -\frac{3}{2}\]Next, substitute \( x = -\frac{3}{2} \) back into the function to get the \( y \)-value:\[y = 2(-\frac{3}{2})^2 + 6(-\frac{3}{2}) = -\frac{9}{2}\]Thus, the vertex is located at \((-\frac{3}{2}, -\frac{9}{2})\). This point is especially crucial for sketching because the parabola will have its 鈥榯urnaround鈥� point here.
X-intercepts
X-intercepts are the points where the graph of the function crosses the x-axis. To find these for any function, set \( f(x) = 0 \) and solve for \( x \).With our example, \( f(x) = 2x^2 + 6x \), setting the equation to zero gives:\[2x^2 + 6x = 0\]Factoring out the common term, we have:\[2x(x + 3) = 0\]Setting each factor to zero:
  • \( 2x = 0 \) gives \( x = 0 \)
  • \( x + 3 = 0 \) gives \( x = -3 \)
Thus, the x-intercepts are \((0, 0)\) and \((-3, 0)\). These points are where the parabola touches or crosses the x-axis, playing a role in shaping the graph.
Y-intercept
The y-intercept is another important point. It鈥檚 where the parabola crosses the y-axis. Finding the y-intercept involves setting \( x = 0 \) in the function.For \( f(x) = 2x^2 + 6x \), substitute \( x = 0 \):\[y = 2(0)^2 + 6(0) = 0\]This means that the y-intercept is at \((0, 0)\). Interestingly, in this case, the y-intercept and one of the x-intercepts are the same point, which makes interpreting the graph a bit simpler. The y-intercept gives you an idea of where the parabola begins as it moves along the x-axis.

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

Recall that the symbol \(\bar{z}\) represents the complex conjugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. $$(\bar{z})^{2}=\overline{z^{2}}$$

Find all solutions of the equation and express them in the form \(a+b i\) $$9 x^{2}+4=0$$

Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$r(x)=\frac{4 x^{2}}{x^{2}-2 x-3}$$

Recall that the symbol \(\bar{z}\) represents the complex conjugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. \(z+\bar{z}\) is a real number.

Evaluate the radical expression and express the result in the form \(a+b i\) $$\frac{1-\sqrt{-1}}{1+\sqrt{-1}}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and 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.