Problem 64
Question
Each equation defines a parabola. Without actually graphing, match the equation in Column I with its description in Column II. A. Vertex \((2,-4) ;\) opens downward B. Vertex \((2,-4) ;\) opens upward C. Vertex \((4,-2) ;\) opens downward D. Vertex \((4,-2) ;\) opens upward E. Vertex \((-2,4) ;\) opens left F. Vertex \((-2,4)\); opens right G. Vertex \((-4,2) ;\) opens left H. Vertex \((-4,2) ;\) opens right $$x=-(y-2)^{2}-4$$
Step-by-Step Solution
Verified Answer
G. Vertex \((-4,2)\); opens left.
1Step 1: Understand Parabola Equation
The given equation is \( x = -(y - 2)^2 - 4 \). This is in the form \( x = a(y - k)^2 + h \), which describes a parabola that opens horizontally. The key features of this equation include the vertex and the direction in which the parabola opens.
2Step 2: Identify the Vertex
From the equation \( x = -(y - 2)^2 - 4 \), we identify the vertex as \( (h, k) \). Here, \( h = -4 \) and \( k = 2 \). Therefore, the vertex is \( (-4, 2) \).
3Step 3: Determine the Direction of Opening
In the equation \( x = a(y - k)^2 + h \), if \( a < 0 \), the parabola opens to the left. Since \( a = -1 \) here (the coefficient of \((y - 2)^2\) is negative), the parabola opens to the left.
4Step 4: Match with Description
The features identified correspond to the description of 'Vertex \((-4, 2)\); opens left' from Column II.
Key Concepts
VertexDirection of OpeningQuadratic Equations
Vertex
The vertex of a parabola is a significant point that often represents either the highest or lowest point of the curve, depending on its orientation. It is essential in identifying and understanding the shape and position of the parabola without the need to graph it. In the context of the given equation, which is expressed as \( x = a(y - k)^2 + h \), the vertex formula reveals that the vertex is located at \((h, k)\). By substituting the values from the equation \( x = -(y - 2)^2 - 4 \), we pinpoint the vertex as \((-4, 2)\). This is critical because it demonstrates the parabola's fixed center of symmetry.
- The first value \( h = -4 \) is the horizontal component, indicating how far left or right from the origin the vertex is placed.
- The second value \( k = 2 \) is the vertical component, showing how far up or down the vertex is positioned.
Direction of Opening
The direction in which a parabola opens is vital for understanding its layout and behavior on a coordinate plane. This direction is determined by the sign of the coefficient \( a \) in the parabola's equation, particularly in forms like \( x = a(y - k)^2 + h \) for horizontal openings. In our particular equation \( x = -(y - 2)^2 - 4 \), the coefficient \( a \) is \(-1\).
Because \( a \) is negative:
Because \( a \) is negative:
- The parabola opens to the left on the coordinate plane.
Quadratic Equations
Quadratic equations are fundamental in describing parabolas, offering a mathematical foundation for understanding their movements and properties. These equations often appear in the form \( y = ax^2 + bx + c \) for vertical parabolas or \( x = a(y - k)^2 + h \) for horizontal parabolas, as in the exercise problem. Here, our focus is on the horizontal parabolas which are less common but equally important.
When given an equation like \( x = -(y - 2)^2 - 4 \), it falls into the category of a quadratic equation with a particular emphasis on expressing \( x \) as a function of \( y \). This orientation results in a curve that extends sideways, highlighting the versatility of quadratic expressions in defining various shapes.
When given an equation like \( x = -(y - 2)^2 - 4 \), it falls into the category of a quadratic equation with a particular emphasis on expressing \( x \) as a function of \( y \). This orientation results in a curve that extends sideways, highlighting the versatility of quadratic expressions in defining various shapes.
- The quadratic component \((y - k)^2\) embodies the curvature.
- The coefficient \( a \) controls the width and orientation of the parabola (whether it narrows or widens as it opens).
Other exercises in this chapter
Problem 63
Each equation defines a parabola. Without actually graphing, match the equation in Column I with its description in Column II. A. Vertex \((2,-4) ;\) opens down
View solution Problem 64
Find an equation for each hyperbola. \(y\) -intercepts \((0, \pm 5) ;\) foci \((0, \pm 3 \sqrt{3})\)
View solution Problem 65
Find an equation for each hyperbola. Asymptotes \(y=\pm \frac{3}{5} x ; y\) -intercepts \((0, \pm 3)\)
View solution Problem 65
For the graph of \((x-h)^{2}=4 c(y-k)\), in what quadrant is the vertex for each condition? (a) \(h0\) (c) \(h>0, k0, k>0\)
View solution