Problem 38
Question
Identify whether each equation, when graphed, will be a parabola, circle,ellipse, or hyperbola. Sketch the graph of each equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. $$ x=-y^{2}+6 y $$
Step-by-Step Solution
Verified Answer
The graph represents a parabola with vertex (9, 3).
1Step 1: Rearrange the equation
The given equation is \(x = -y^{2} + 6y\). We need to rearrange it in a standard form commonly used for identifying conic sections. Rearrange it as \( x = -(y^{2} - 6y) \).
2Step 2: Complete the square
To rewrite the equation in a recognizable form, complete the square for the terms involving \( y \). Start with \( y^{2} - 6y \). The number needed to complete the square is \((\frac{6}{2})^2 = 9\). Add and subtract 9 inside the square bracket: \( x = -(y^{2} - 6y + 9 - 9) \).
3Step 3: Simplify the equation
Simplifying the expanded terms gives us \( x = -( (y - 3)^2 - 9 ) \), which further simplifies to \( x = -(y - 3)^2 + 9 \).
4Step 4: Identify the conic section
The equation \( x = -(y - 3)^2 + 9 \) is in the form \( x = a(y - k)^2 + h \), which is the standard form of a vertical parabola. Hence, this equation represents a parabola.
5Step 5: Identify and label key features of the parabola
The vertex of the parabola is found at \( (h, k) = (9, 3) \). As the parabola is in the form of \( x = -(y - 3)^2 + 9 \), it opens to the left because of the negative coefficient.
6Step 6: Sketch the graph
Sketch a parabola opening to the left with the vertex at the point \( (9, 3) \). The axis of symmetry is the line \( y = 3 \).
Key Concepts
ParabolaCompleting the SquareVertex
Parabola
A parabola is a U-shaped curve that can open upward, downward, left, or right. In mathematics, it is a particular type of conic section. When you see an equation involving a squared term, such as in this exercise, it often indicates a parabola.- A parabola can be expressed in the standard form * - If it is vertical: \( y = a(x - h)^2 + k \) *- If it is horizontal: \( x = a(y - k)^2 + h \) The equation given in the problem, \( x = -(y - 3)^2 + 9 \), fits the horizontal parabola format. Here, the squared term is with \( y \), suggesting that it opens along the x-axis.Understanding the direction in which the parabola opens is crucial. In this case, because of the negative sign before the squared term, the parabola opens to the left. This character is typical for such arrangements when the equation has a negative leading coefficient.
Completing the Square
Completing the square is a method used to transform quadratic expressions into a form that is easier to work with, typically into perfect square trinomials. This process is invaluable in understanding the conic sections, especially parabolas.To "complete the square" for a quadratic equation, such as \(y^2 - 6y\), follow these steps:- Identify the coefficient of the linear term, which is \(-6\).- Divide this coefficient by 2, and then square the result: * \((\frac{-6}{2})^2 = 9\).- Add and subtract this square inside the equation: * \(y^2 - 6y + 9 - 9\).Reorganize the terms to form a perfect square trinomial \((y - 3)^2\), which rearranges the problem into a familiar format. This is because \((y - 3)^2\) expands to \(y^2 - 6y + 9\).Completing the square allows us to directly see the shifts of the parabola, leading us neatly to identifying the vertex without extra guesswork.
Vertex
The vertex of a parabola is the point where it changes direction. In simpler terms, it's the peak or the lowest point of the U-shape, depending on how the parabola opens.For the given horizontal parabola equation \(x = -(y - 3)^2 + 9\), the vertex is at the point \((h, k)\). Here, this is clearly seen as \((9, 3)\).- The vertex - Provides a point of symmetry for the parabola, meaning the curve is mirrored on both sides around this point.- Helps determine the direction in which the parabola opens. A positive \( a \) value would open rightwards for horizontal parabolas, while a negative one, as here, results in a leftward opening.Graphically, having the vertex helps you plot the parabola accurately and understanding its geometrical nature gives insight into its real-world applications, such as in satellite dishes or reflecting telescopes, where the vertex plays a key role in focusing signals or light.
Other exercises in this chapter
Problem 37
Hint: For Exercises 33 through 38 , first divide the equation through by the coefficient of \(x^{2}\) (or \(\left.y^{2}\right)\). $$ 4(x+1)^{2}+4(y-3)^{2}=12 $$
View solution Problem 37
Without graphing, how can you tell that the graphs of \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\) do not have any points of intersection?
View solution Problem 38
Hint: For Exercises 33 through 38 , first divide the equation through by the coefficient of \(x^{2}\) (or \(\left.y^{2}\right)\). $$ 5(x-2)^{2}+5(y+1)=50 $$
View solution Problem 38
Without solving, how can you tell that the graphs of \(y=2 x+3\) and \(y=2 x+7\) do not have any points of intersection?
View solution