/*! 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 83 (a)write the equation in standar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 83

(a)write the equation in standard form and (b) use properties of the standard form to graph the equation. $$ x=-2 y^{2}-12 y-16 $$

Short Answer

Expert verified
The standard form is \[ x = -2(y + 3)^2 + 2 \]. The vertex is \((-3, 2)\) and the parabola opens left.

Step by step solution

01

- Rearrange the equation

Start by rearranging the equation to isolate the quadratic terms on one side. The given equation is: \[ x = -2y^2 - 12y - 16 \]Let's rewrite this equation to be in the form of \[ x = A(y + h)^2 + k \]
02

- Complete the square

Complete the square for the quadratic expression in y. Begin with the quadratic term and linear term: \[ x = -2(y^2 + 6y) - 16 \]Add and subtract the square of half the coefficient of y, \[ y^2 + 6y + 9 - 9 = (y + 3)^2 - 9 \]Thus, the equation becomes: \[ x = -2((y + 3)^2 - 9) - 16 \]
03

- Simplify the equation

Distribute the -2 and simplify the equation:\[ x = -2(y + 3)^2 + 18 - 16 \]Combine constants to get the equation in standard form\[ x = -2(y + 3)^2 + 2 \]
04

- Determine vertex and direction

In standard form, the equation is \[ x = A(y + h)^2 + k \]where the vertex (h, k) is the turning point of the parabola. For \[ x = -2(y + 3)^2 + 2 \]the vertex is \( (-3, 2) \) and the parabola opens horizontally towards the left since A is negative.
05

- Plot the vertex

Plot the vertex on the coordinate plane at the point \((-3, 2)\).
06

- Find and plot additional points

To better understand the shape of the parabola, find additional points: For example, when \( y = 0 \) then \[ x = -2(0 + 3)^2 + 2 = -2(9) + 2 = -18 + 2 = -16 \]Plot this point \((-16, 0)\). Similarly find other points and plot them.
07

- Draw the parabola

Using the vertex and other points, draw a smooth curve to form the parabola opening to the left.

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 Quadratic Equations
The standard form of a quadratic equation for parabolas is a helpful format for graphing and understanding the curve. It is generally written as:

\ x = A(y + h)^2 + k

Here:
  • \(A\) is a constant that influences the direction and the width of the parabola. A negative \(A\) means the parabola opens leftwards; a positive \(A\) means it opens rightwards.
  • \(h\) and \(k\) influence the location of the vertex of the parabola.
In our exercise, we started with the equation: \[ x = -2y^2 - 12y - 16 \]
This needed to be transformed into the standard form by completing the square, which simplifies identifying features like the vertex and direction.
Completing the Square
Completing the square is a method used to convert a quadratic equation into its standard form. It's especially useful for graphing parabolas. Here's how it works in this exercise:

We begin with the equation:

\[ x = -2(y^2 + 6y) - 16 \]
To complete the square, we add and subtract the same value inside the brackets to form a perfect square trinomial:

\[ y^2 + 6y + 9 - 9 = (y + 3)^2 - 9 \]
Now our equation becomes:

\[ x = -2((y + 3)^2 - 9) - 16 \]
Finally, we simplify it:

\[ x = -2(y + 3)^2 + 18 - 16 \]
\[ x = -2(y + 3)^2 + 2 \]
This is the standard form, making it easier to identify the properties of the parabola for graphing.
Graphing Parabolas
Graphing a parabola involves plotting key features derived from its standard form equation. Let's apply this to our example

\[ x = -2(y + 3)^2 + 2 \]

First, identify the vertex:
  • From \((y + h)\), we find \(h = -3\)
  • From \(k\), we find \(k = 2\)
  • Therefore, the vertex is at \((-3, 2)\)
Since the coefficient \(A = -2\) is negative, the parabola opens towards the left.

Next, plot the vertex:

  • Draw a point at \((-3, 2)\)
To shape the parabola, find additional points by substituting values of \(y\). Example:

When \( y = 0 \):

\[ x = -2(0 + 3)^2 + 2 \]
\[ x = -2(9) + 2 \]
\[ x = -18 + 2 \]
\[ x = -16 \]

Plot this point at \((-16, 0)\).

Repeat for more points, then draw a smooth curve through these points and the vertex to complete the graph of the parabola.

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

Graph. $$ \frac{(x-2)^{2}}{4}-\frac{(y-3)^{2}}{16}=1 $$

Graph. $$ \frac{(y-1)^{2}}{25}-\frac{(x-4)^{2}}{16}=1 $$

Solve the system of equations by using substitution. $$ \left\\{\begin{array}{l} 9 x^{2}+4 y^{2}=36 \\ x=2 \end{array}\right. $$

Solve the system of equations by using substitution. $$ \left\\{\begin{array}{l} 2 y^{2}-x=0 \\ y=x+1 \end{array}\right. $$

Solve the system of equations by using elimination. $$ \left\\{\begin{array}{l} 4 x^{2}-y^{2}=4 \\ 4 x^{2}+y^{2}=4 \end{array}\right. $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

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.