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Chapter 13: Problem 50

Write each equation of a parabola in standard form and graph it. Give the coordinates of the vertex. $$ x=-2 y^{2}+4 y+1 $$

Short Answer

Expert verified
The parabola in standard form is \( x = -2(y-1)^2 + 3 \) with vertex at (3, 1).

Step by step solution

01

Rearrange the Equation

Start by rearranging the given equation, which is originally in a non-standard form for a parabola. The equation is given as \[ x = -2y^2 + 4y + 1 \]Notice that this is a quadratic in terms of \( y \), so we want to complete the square for the terms containing \( y \).
02

Factor Out the Coefficient of the Quadratic Term

Factor out the coefficient of \( y^2 \) from the terms with \( y \):\[ x = -2(y^2 - 2y) + 1 \]
03

Complete the Square

To complete the square inside the parentheses:1. Take half of the coefficient of \( y \), which is \( -2 \), giving \( -1 \).2. Square \( -1 \) to get \( 1 \). 3. Add and subtract \( 1 \) inside the parentheses: \[ x = -2(y^2 - 2y + 1 - 1) + 1 \] 4. Rearrange to create a perfect square trinomial: \[ x = -2((y-1)^2 - 1) + 1 \]
04

Simplify the Equation

Distribute the \( -2 \) and simplify:\[ x = -2(y-1)^2 + 2 + 1 \] \[ x = -2(y-1)^2 + 3 \]Now the parabola is in standard form, showing a horizontal parabola with vertex form \( x = a(y-k)^2 + h \).
05

Identify the Vertex

From the equation, the vertex form is \((x = a(y-k)^2 + h)\). The vertex \((h, k)\) is \( (3, 1) \). Thus, the coordinates of the vertex for this parabola are \( (3, 1) \).
06

Graph the Parabola

The parabola \( x = -2(y-1)^2 + 3 \) opens to the left because the coefficient \( a = -2 \) is negative.1. Plot the vertex \( (3, 1) \) on the coordinate plane.2. Since the parabola opens horizontally, choose a few values for \( y \) around \( y = 1 \) to find corresponding \( x \) values.3. Sketch the curve based on the plotted points and ensure it is symmetric around the line \( y = 1 \).The graph will show a leftward opening parabola with the vertex at \( (3, 1) \).

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Key Concepts

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

Vertex Form
The vertex form of a parabola is a way of expressing the quadratic equation that highlights the vertex, making it easier to graph. The generic vertex form for a parabola is either \( y = a(x-h)^2 + k \) for a parabola that opens upward or downward, or \( x = a(y-k)^2 + h \) for a parabola that opens sideways.
  • Here, \( (h, k) \) represent the vertex of the parabola.
  • The value of \( a \) affects the direction and the width of the parabola.
In our given equation \( x = -2(y-1)^2 + 3 \), the vertex form immediately tells us that the vertex of the parabola is located at the point \((3, 1)\). Additionally, since \( a = -2 \), it indicates that this parabola opens to the left, due to the negative value.
Completing the Square
Completing the square is a method used to transform a quadratic equation in a way that helps identify key features of a parabola, like its vertex. The method involves rewriting a quadratic equation such that it includes a perfect square trinomial.
  • Start by factoring out any coefficient in front of the quadratic term.
  • Take half of the linear coefficient, square it, and then add and subtract this square within the equation to complete the square.
  • This creates a perfect square trinomial inside the brackets which can be easily rewritten as \((variable - constant)^2\).
For example, in the original equation \( x = -2y^2 + 4y + 1 \), factor out \(-2\) from the \(y^2\) and \(y\) terms, take half of \(-2\) (which is \(-1\)), square it to get \(1\), and complete the square:\[ x = -2(y^2 - 2y + 1 - 1) + 1 = -2((y-1)^2 - 1) + 1 \] This procedure helps transform the expression into a form that reveals the vertex and makes graphing easier.
Graphing Parabolas
Graphing a parabola can be straightforward once you have it in vertex form. Here鈥檚 how you can effectively graph it:
  • First, plot the vertex of the parabola, which serves as a key reference point.
  • Determine the direction of opening: a positive \(a\) value indicates it opens right or upwards, while a negative \(a\) indicates it opens left or downwards. In this particular case, since \(a = -2\), the parabola opens to the left.
  • Find additional points by choosing values for the variable and solving for the other variable.
  • Since a parabola is symmetric, use symmetry to ensure points reflect around the axis of symmetry, which in this example is \(y = 1\).
By following these steps with our example \( x = -2(y-1)^2 + 3 \), we can see that the vertex \((3, 1)\) positions the graph with its symmetrical axis at \(y = 1\), leading the parabola to curve leftwards.

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Most popular questions from this chapter

Write each equation in standard form, if it is not alreacty so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ x^{2}+y^{2}+2 x-8=0 $$

Solve each system of equations for real values of x and y. $$ \left\\{\begin{array}{l} 2 x^{2}-3 y^{2}=5 \\ 3 x^{2}+4 y^{2}=16 \end{array}\right. $$

Compare the graphs of \(\frac{x^{2}}{81}+\frac{y^{2}}{64}=1\) and \(\frac{x^{2}}{64}+\frac{y^{2}}{81}=1 .\) Do they have any similarities?

Graph: \(9 x^{2}+4 y^{2}=1\)

APPLICATIONS Radio stations applying for licensing may not use the same frequency if their broadcast areas overlap. One station's coverage is bounded by \(x^{2}+y^{2}-8 x-20 y+16=0,\) and the other's by \(x^{2}+y^{2}+2 x+4 y-11=0 .\) May they be licensed for the same frequency?
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