Chapter 12: Problem 28
Sketch the graph of each equation. $$ 20 y^{2}-40 y-x=-25 $$
Short Answer
Expert verified
The graph is a rightward-opening parabola with vertex at (5,1), focus at (10,1), and directrix at x=0.
Step by step solution
01
Rewrite equation in standard form
First rewrite the equation in standard form to see if it's a parabola with the vertex not at the origin. Standard form for a parabola is \( (y - k)^2 = 4p (x - h) \). Here, \( 20y^{2}-40y-x=-25 \) is the given equation.
02
Solve for x
Next, isolate x. The equation becomes \( x = 20y^2 - 40y + 25 \).
03
Complete the Square
Now, complete the square on the right-hand side by adding and subtracting the proper term. Adding and subtracting 4 gives \( x = 20(y - 1) ^ 2 - 5 \).
04
Identify Vertex, Focus and Directrix
Now we can see this is a parabola opening to the right. The vertex of the parabola is \( (h, k) = (5, 1) \). The value of 4p is 20 so p = 5. The focus is \( (h + p, k) = (10, 1) \) and the directrix is \( x = h - p = 0 \).
05
Sketch the Graph
Using the vertex, the focus and the directrix, sketch the graph. This is a rightward-opening parabola with the vertex at (5,1), the focus at(10,1), and the directrix given by the vertical line x=0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a method used to transform any quadratic equation into a perfect square trinomial. This technique helps simplify the process of solving for variables, and is particularly useful in turning equations into a form that makes graphing parabolas easier. Here's how it generally works:
- First, ensure the quadratic and linear terms are on one side of the equation. In this example, we rearranged the terms as \( x = 20y^2 - 40y + 25 \).
- Next, factor out the coefficient of the quadratic term (here, 20) from the terms involving \( y \). This gives us:\( x = 20(y^2 - 2y) + 25 \).
- Find the value needed to complete the square: Take half of the linear coefficient (\(-2\)), square it (\((-2/2)^2 = 1\)), and add and subtract it within the parentheses: \( x = 20((y^2 - 2y + 1) - 1) + 25 \).
- Simplify to the completed square form:\( x = 20(y - 1)^2 - 5 \).
Vertex of a Parabola
The vertex of a parabola is a key component that represents the turning point of the graph. It is either the highest or lowest point, depending on the orientation of the parabola.In the standard form equation \((y - k)^2 = 4p(x - h)\), the vertex of the parabola is directly found at the point \((h, k)\).Given the equation after completing the square, \( x = 20(y - 1)^2 - 5 \),we can see:
- \( h = 5 \)
- \( k = 1 \)
Focus and Directrix
The focus and directrix are fundamental elements in defining the shape and orientation of a parabola. These components are linked by the concept that a parabola consists of all points equidistant from a point (focus) and a line (directrix).Once we have our parabolic equation in the standard form, we identify the location of these elements:
- The given completed square equation, \( x = 20(y - 1)^2 - 5 \), can be rewritten as \((y - 1)^2 = 4 \cdot 5 \cdot (x - 5)\),indicating \( 4p = 20 \), leading us to \( p = 5 \).
- The focus is found at \((h + p, k) = (10, 1)\), meaning it is located to the right of the vertex, indicating a rightward opening parabola.
- The directrix, a vertical line at \( x = h - p = 0 \), is positioned to the left of the vertex.
Equation of a Parabola
The equation of a parabola can take various forms, but commonly it is expressed in the vertex form as \((y - k)^2 = 4p(x - h)\). This format highlights the significant features of a parabola, such as its vertex, focus, and directrix, guiding us in graphing and analyzing the curve.In the given exercise, we start with a quadratic equation and through completing the square and adjusting terms, it is rewritten into a form that easily gives insight into the parabola's key features.
- The original equation \( x = 20y^2 - 40y + 25 \),when rewritten as \((y - 1)^2 = 4 \cdot 5 \cdot (x - 5)\), signifies a parabolic graph where:
- \( h = 5 \) and \( k = 1 \), giving the vertex \((5, 1)\).
- \( p = 5 \), confirming the primary distance used in locating the focus and directrix.