Problem 43
Question
Graph each equation of a parabola. Give the coordinates of the vertex. $$ x=y^{2} $$
Step-by-Step Solution
Verified Answer
The parabola opens to the right, with the vertex at (0, 0) and axis of symmetry along the x-axis.
1Step 1: Understand the Parabolic Equation
The given equation is \( x = y^2 \). Typically, equations are given in the form \( y = x^2 \), but here, the roles of \( x \) and \( y \) are switched, indicating that the parabola opens horizontally.
2Step 2: Rewrite the Equation in a More Recognizable Vertex Form
For a horizontal parabola, the vertex form is \( x = a(y - k)^2 + h \). Here, we can see that \( x = (y - 0)^2 + 0 \), meaning it is already in vertex form with \( h = 0 \) and \( k = 0 \).
3Step 3: Identify the Vertex of the Parabola
From the vertex form \( x = (y - 0)^2 + 0 \), we identify the vertex as the point \( (h, k) \). Substituting the values of \( h \) and \( k \), the vertex is \( (0, 0) \).
4Step 4: Sketch the Parabola Using Key Points
Since the parabola opens to the right, use key points to sketch it. Calculate points such as when \( y = 1 \) (\( x = 1 \)), \( y = 2 \) (\( x = 4 \)), and \( y = -1 \) (\( x = 1 \)), \( y = -2 \) (\( x = 4 \)). Plot these points on a graph and draw the curve showing that the parabola opens to the right.
Key Concepts
Vertex of a ParabolaHorizontal ParabolaParabola Sketching
Vertex of a Parabola
In any parabola, whether it opens upwards, downwards, or horizontally, the vertex is a crucial point that gives us insight into the parabola's geometry. The vertex is essentially the point where the parabola changes direction. For vertical parabolas, it's either the lowest or highest point. For horizontal parabolas, like in our equation \(x = y^2\), it is the leftmost or rightmost point. This point provides crucial information about the graph.
In our equation \(x = (y - 0)^2 + 0\), the vertex form \(x = a(y - k)^2 + h\) reveals the vertex directly from \(h\) and \(k\), which correspond to the \(x\)- and \(y\)-coordinates. Here, both \(h\) and \(k\) are 0, indicating that the vertex of this parabola is at the origin, \((0, 0)\). This simple location helps in plotting and understanding the orientation and symmetry of the parabola.
In our equation \(x = (y - 0)^2 + 0\), the vertex form \(x = a(y - k)^2 + h\) reveals the vertex directly from \(h\) and \(k\), which correspond to the \(x\)- and \(y\)-coordinates. Here, both \(h\) and \(k\) are 0, indicating that the vertex of this parabola is at the origin, \((0, 0)\). This simple location helps in plotting and understanding the orientation and symmetry of the parabola.
Horizontal Parabola
A horizontal parabola opens to the side instead of vertically like the more traditional parabolas. The standard form of these is helpful to determine their direction and orientation quickly. Whereas the typical vertical parabolas follow \(y = ax^2 + bx + c\), horizontal parabolas have equations shaped like \(x = a(y-k)^2 + h\).
In this form, the direction in which the parabola opens (left or right) is determined by the sign of the coefficient \(a\):
In this form, the direction in which the parabola opens (left or right) is determined by the sign of the coefficient \(a\):
- If \(a > 0\), the parabola opens to the right.
- If \(a < 0\), it opens to the left.
Parabola Sketching
Sketching a parabola, especially when it is horizontal, involves understanding its structure and key points which help in drawing it accurately. The first step is identifying the vertex, which acts as an anchor point.
To graph \(x = y^2\), begin with the identified vertex \((0, 0)\). This parabola opens to the right due to its positive \(a\) value. Now plot key points such as:
To graph \(x = y^2\), begin with the identified vertex \((0, 0)\). This parabola opens to the right due to its positive \(a\) value. Now plot key points such as:
- When \(y = 1\), then \(x = 1\) so point is \((1, 1)\).
- When \(y = 2\), then \(x = 4\) so point is \((4, 2)\).
- When \(y = -1\), then \(x = 1\) so point is \((1, -1)\).
- When \(y = -2\), then \(x = 4\) so point is \((4, -2)\).
Other exercises in this chapter
Problem 43
Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y=x^{2}-4 \\ x^{2}-y^{2}=-16 \end{array}\right. $$
View solution Problem 43
Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hype
View solution Problem 44
Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y-x=0 \\ 4 x^{2}+y^{2}=10 \end{array}\right. $$
View solution Problem 44
Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hype
View solution