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Chapter 13: Problem 44

Graph each equation of a parabola. Give the coordinates of the vertex. $$ x=2 y^{2} $$

Short Answer

Expert verified
The vertex of the parabola is at (0, 0), and it opens to the right.

Step by step solution

01

Identify the Standard Form

The given equation is \( x = 2y^2 \). This is a sideways parabola, different from the commonly seen \( y = ax^2 \). Its standard form is \( x = a(y-k)^2 + h \), where \((h, k)\) is the vertex of the parabola.
02

Complete the Equation

The equation \( x = 2y^2 \) is already in the form \( x = 2(y-0)^2 + 0 \). Here, \(h = 0\) and \(k = 0\), hence \((h, k) = (0,0)\).
03

Identify the Vertex

From the standard form identified in Step 2, the vertex of the parabola \( x = 2y^2 \) is \((h, k) = (0, 0)\).
04

Sketch the Graph of the Parabola

The graph of the equation \( x = 2y^2 \) is a parabola opening to the right because the coefficient of \( y^2 \), which is 2, is positive. Plot points like \((-2, -1), (0, 0), (2, 1)\) to sketch the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex of a Parabola
In a parabola, the vertex is an important feature as it represents the point where the curve changes direction. It can be a maximum or minimum point, but in the case of sideways parabolas, like the equation \(x = 2y^2\), it's where the graph shifts from opening left to right.
  • In standard parabolic equations for vertical orientation, \(y = ax^2\), the vertex is often at the peak or the trough of the curve.
  • For sideways parabolas, as in \(x = a(y-k)^2 + h\), the vertex is still central but affects horizontal movement.
To identify the vertex from an equation such as \(x = 2(y-0)^2 + 0\), we look for values of \(h\) and \(k\). Here, both are zero, making the vertex \((0, 0)\). Knowing the vertex helps with sketching and understanding the graph's orientation.
Standard Form of a Parabola
The standard form of a parabola provides a blueprint for its graph. For vertical parabolas, the form is usually \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex. However, when dealing with horizontal parabolas, as seen here, it's \(x = a(y-k)^2 + h\).
  • The coefficient \(a\) defines the parabola鈥檚 width and direction. If \(a\) is positive, the graph opens upwards for vertical or rightwards for horizontal.
  • When \(a\) is negative, the graph inverts, opening downward or leftward.
For \(x = 2y^2\), rewriting it in standard form shows \(a = 2\), and since \(h = 0\) and \(k = 0\), it simplifies our understanding. Identifying these components is crucial as they shape how we graph the equation, locate the vertex, and analyze the parabola's characteristics.
Graphing Parabolas
Graphing a parabola involves plotting points to reflect its curve and opening direction. Start by identifying the vertex using the standard form components \((h, k)\) and plotting that first.For \(x = 2y^2\), where the parabola opens to the right:
  • Once the vertex \( (0, 0) \) is placed on the graph, choose additional points easily offset from the vertex to get a sense of the curvature, like \((-2, -1), (0, 0), (2, 1)\).
  • The positive \(a\) value of 2 indicates that as \(y\) moves away from zero, \(x\) increases in the direction the parabola opens (rightward in this case).
  • The symmetry about the vertex also helps. For each point \((y, x)\), consider its symmetric counterpart, \((-y, x)\).
Plotting these allows us to sketch an accurate curve. Ensure to also check the directionality confirmed through the vertex found, here revealing a sideways opening parabola opposed to the usual upward/downward shape.

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