Chapter 13: Problem 80
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. $$ x=-\frac{1}{3} y^{2}-2 y $$
Short Answer
Expert verified
The equation represents a parabola with vertex at (3, -3) and opens to the left.
Step by step solution
01
Identify the type of equation
The given equation is \( x = -\frac{1}{3} y^2 - 2y \). This equation is in the form \( x = ay^2 + by + c \), which is characteristic of a parabola that opens either left or right. This is not a circle since it has an equation involving \( y^2 \) but no \( x^2 \) term.
02
Rewrite the equation in standard form
The standard form of a parabola that opens sideways (left or right) is \( x = a(y-k)^2 + h \) where \((h, k)\) is the vertex. To rewrite the equation \( x = -\frac{1}{3} y^2 - 2y \) in standard form, we need to complete the square on the \( y \) terms.
03
Complete the square
Start by factoring out the coefficient of \( y^2 \) from the terms involving \( y \): \[ x = -\frac{1}{3} (y^2 + 6y) \].Next, complete the square for the expression \( y^2 + 6y \). Take half of 6, square it to get 9, and add and subtract 9 inside the bracket: \[ x = -\frac{1}{3} (y^2 + 6y + 9 - 9) \].Simplifying inside the brackets gives: \[ x = -\frac{1}{3} ((y + 3)^2 - 9) \].
04
Simplify the expression
Distribute the \(-\frac{1}{3}\) into the completed square expression: \[ x = -\frac{1}{3}(y + 3)^2 + 3 \].Now, the equation is in the standard form \( x = a(y-k)^2 + h \) with \( a = -\frac{1}{3} \), \( h = 3 \), and \( k = -3 \).
05
Identify the vertex
From the equation \( x = -\frac{1}{3}(y + 3)^2 + 3 \), the vertex of the parabola is at \((h, k) = (3, -3)\).
06
Determine the direction of the parabola's opening
Since \( a = -\frac{1}{3} \) is negative, the parabola opens to the left. This is because a negative coefficient before the squared term indicates a leftward opening for parabolas in the form \( x = a(y-k)^2 + h \).
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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 technique used to transform a quadratic equation into a form that easily reveals important features of the graph, like the vertex of a parabola. This technique is especially useful when the equation is initially not in an easily interpretable form.
Here's how you typically go about completing the square:
- Start by identifying the quadratic and linear terms. In the context of the original equation, this was the part involving y, specifically ²â² and 2y (or 6y after factoring).
- Then, you'll take half of the coefficient of the linear term (in our case, half of 6 is 3), and square it (3² = 9).
- Add and subtract this square inside the equation to form a perfect square trinomial (like the step
²â² + 6y + 9). - This trinomial is rewritten as a square of a binomial, giving us (y + 3)². That way, we have a simplified form.
Standard Form of a Parabola
The standard form of a parabola is key in determining the orientation and position of the graph in a straightforward manner. For parabolas that open left or right, which involve terms like ²â², the standard form is:\[ x = a(y-k)² + h \]Here,
- (h, k) represents the vertex, the point where the parabola is most curved.
- a influences the width and direction in which the parabola opens.
Equation of a Parabola
The equation of a parabola provides all the necessary information to graph it. When given a parabolic equation, our goal is often to identify key features like:
- The position of the vertex: Found in the standard form, this is critical for graphing as it acts as the 'starting point' for plotting.
- Directionality: Figured out through the a value, it determines whether your graph opens upwards, downwards, left, or right.