Problem 65
Question
For the graph of \((x-h)^{2}=4 c(y-k)\), in what quadrant is the vertex for each condition? (a) \(h<0, k<0\) (b) \(h<0, k>0\) (c) \(h>0, k<0\) (d) \(h>0, k>0\)
Step-by-Step Solution
Verified Answer
(a) Quadrant III, (b) Quadrant II, (c) Quadrant IV, (d) Quadrant I.
1Step 1: Understanding the Equation
The equation \((x-h)^2 = 4c(y-k)\) describes a parabola. In this form, \( (h, k) \) represents the vertex of this parabola.
2Step 2: Identifying the Quadrants
Recall that in the Cartesian plane, the four quadrants are defined as follows: 1. Quadrant I: Both \(x\) and \(y\) are positive. 2. Quadrant II: \(x\) is negative, \(y\) is positive. 3. Quadrant III: Both \(x\) and \(y\) are negative. 4. Quadrant IV: \(x\) is positive, \(y\) is negative.
3Step 3: Condition (a): h<0, k<0
Given \( h<0 \) and \( k<0 \), the vertex \((h, k)\) has both a negative \(x\)-coordinate and a negative \(y\)-coordinate, which places it in Quadrant III.
4Step 4: Condition (b): h 0
For \( h<0 \) and \( k>0 \), the vertex \((h, k)\) has a negative \(x\)-coordinate and a positive \(y\)-coordinate, situating it in Quadrant II.
5Step 5: Condition (c): h>0, k<0
Here, \( h>0 \) and \( k<0 \) make the vertex \((h, k)\) have a positive \(x\)-coordinate and a negative \(y\)-coordinate, placing it in Quadrant IV.
6Step 6: Condition (d): h>0, k>0
With \( h>0 \) and \( k>0 \), both coordinates of the vertex \((h, k)\) are positive, locating it in Quadrant I.
Key Concepts
Vertex of a ParabolaUnderstanding QuadrantsIntroduction to the Cartesian Plane
Vertex of a Parabola
The vertex is a crucial point on a parabola that determines its position in the Cartesian plane. In the equation \((x-h)^2 = 4c(y-k)\), the vertex is given by the coordinates \((h, k)\). It represents the point where the parabola changes direction. Whether the curve opens upwards or downwards depends on the value of \(c\), but the vertex is always the pivot point for the parabola's shape.
To fully understand the vertex's role, consider some key points:
To fully understand the vertex's role, consider some key points:
- The vertex is the minimum or maximum point of the parabola, depending on its orientation.
- It helps in graphing the parabola accurately by providing a reference point.
- The coordinates \((h, k)\) determine where the vertex is located relative to the origin of the Cartesian plane.
Understanding Quadrants
In mathematics, quadrants are crucial for pinpointing locations on a Cartesian plane. There are four quadrants, and knowing them can help you determine where a point like a vertex lies.
- Quadrant I has both coordinates positive, meaning any point here has \(x > 0\) and \(y > 0\).
- Quadrant II has a negative \(x\) and a positive \(y\), which is where any \((x, y)\) with \(x < 0\) and \(y > 0\) is located.
- Quadrant III is home to negative coordinates on both axes, so \((x, y)\) points here have \(x < 0\) and \(y < 0\).
- Quadrant IV, conversely, has a positive \(x\) and negative \(y\), meaning any \((x, y)\) will be \(x > 0\) and \(y < 0\).
Introduction to the Cartesian Plane
A Cartesian plane is a two-dimensional coordinate system that facilitates plotting mathematical functions and points in a defined space. It consists of horizontal (\(x\)-axis) and vertical (\(y\)-axis) lines that intersect at the origin \((0,0)\). This creates a plane divided into four quadrants which are useful for graphical analysis.
The Cartesian plane serves many purposes:
The Cartesian plane serves many purposes:
- It provides a clear visual representation of algebraic expressions and geometric figures.
- It helps in graphing parabolas, making it easier to analyze their properties such as direction and vertex location.
- By understanding its axes and quadrants, one can determine the position of any point, including where the parabola's vertex falls, helping to solve complex mathematical problems with ease.
Other exercises in this chapter
Problem 64
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
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Find an equation for each hyperbola. Asymptotes \(y=\pm \frac{3}{5} x ; y\) -intercepts \((0, \pm 3)\)
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Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}+y^{2} &=10 \\ 2 x^{2}-y^{2} &=17 \end{aligned}$$
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Find an equation for each hyperbola. \(y\) -intercept \((0,-2)\); center at origin; passing through \((2,3)\)
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