/*! 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 59 Sketch the graph of each equatio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$x=-3 y^{2}+30 y$$

Short Answer

Expert verified
The graph is a parabola with vertex at (75, 5), opening left.

Step by step solution

01

Identify the Type of Graph

The given equation is \[ x = -3y^2 + 30y \]This resembles the standard form of a parabola equation with the form \[ x = ay^2 + by + c \]where \( a = -3 \), \( b = 30 \), and \( c = 0 \). Thus, this graph is a parabola.
02

Complete the Square

To find the vertex, we need to complete the square. Start by rewriting the equation:\[ x = -3(y^2 - 10y) \]Next, complete the square for the expression inside the parentheses:\[ y^2 - 10y \]Here, take half of \(-10\) (which is \(-5\)), square it (\((-5)^2 = 25\)), and add it inside the parentheses:\[ y^2 - 10y = (y - 5)^2 - 25 \]Substitute back into the original equation:\[ x = -3((y - 5)^2 - 25) = -3(y - 5)^2 + 75 \]
03

Identify the Vertex

The vertex form of a parabola is\[ x = a(y-k)^2 + h \]Here, from \[ x = -3(y - 5)^2 + 75 \]we can identify \( h = 75 \) and \( k = 5 \). Therefore, the vertex of the parabola is at the point \((75, 5)\).
04

Sketch the Graph

On a coordinate plane, plot the vertex of the parabola at (75, 5). The coefficient of \((y - 5)^2\) is \(-3\), indicating that the parabola opens to the left because it is a negative value. Sketch the parabola with the vertex (75, 5), noting the leftward opening due to the negative coefficient.

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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 transform a quadratic expression into a perfect square trinomial. This is particularly useful in finding the vertex form of a parabola, which makes it easier to identify key features like the vertex.

For the equation \[x = -3y^2 + 30y\]we'll start by focusing on the part inside the parentheses when \[x = -3(y^2 - 10y)\].
To complete the square:
  • Take half of the linear coefficient, which in this case is -10, giving us -5.
  • Square it: \((-5)^2 = 25\). This is the number we'll use to complete the square.
Rewrite the quadratic expression as a perfect square trinomial:\[y^2 - 10y = (y - 5)^2 - 25\]
This transformation' will help us rewrite the original equation into vertex form, aiding in graphing and analysis.
Vertex of a Parabola
The vertex of a parabola is the point where it turns around; it's the highest or lowest point depending on the parabola’s direction. For our parabola, using the completed square, we rewrite the equation\[x = -3(y - 5)^2 + 75\]This equation is now in vertex form:
  • Here, \((y - 5)^2\) represents the squared term.
  • The number 5 is the y-component of the vertex, \(k = 5\).
  • The term 75 is the x-component of the vertex, \(h = 75\).
This means the vertex of this parabola is at the point \((75, 5)\).
Understanding the vertex helps in plotting the graph of the parabola efficiently.
Equation of a Parabola
The equation of a parabola can be expressed in different forms, and understanding these can greatly simplify the graphing and analysis process. In the context of our exercise, we began with a standard quadratic form:\[x = -3y^2 + 30y\]Translating this into vertex form, as we completed the square, gave us:\[x = -3(y - 5)^2 + 75\]Vertex form is highly beneficial because it clearly outlines the vertex of the parabola. By looking at \[x = a(y-k)^2 + h\], where \(a\) indicates the direction and width of the parabola, and \((h, k)\) represents the vertex.
  • Here, \(a = -3\), indicating the parabola opens to the left because it is negative.
  • The values \(h = 75\) and \(k = 5\)form the vertex.
Recognizing this format helps sketch the parabola accurately and illuminate the behavior of the quadratic function visually on a graph.

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