Chapter 6: Problem 4
Put the equation \(2 y^{2}-x+12 y+16=0\) into standard form and graph the resulting parabola.
Short Answer
Expert verified
The parabola is \((y + 3)^2 = \frac{1}{2}(x + 2)\) with vertex at \((-2, -3)\) opening to the right.
Step by step solution
01
Recognize the Parabola Equation
The equation provided is not in standard parabola form. A parabola in standard form with a horizontal axis can be expressed as \((y - k)^2 = 4p(x - h)\). Our goal is to rewrite the equation \(2y^2 - x + 12y + 16 = 0\) in a similar format.
02
Rearrange Terms
Rearrange the original equation to group the \(y\)-terms together: \(2y^2 + 12y - x + 16 = 0\) becomes \(2y^2 + 12y = x - 16\).
03
Complete the Square for y Terms
To complete the square for the \(y\) terms, factor out the coefficient of \(y^2\) which is 2, giving us \(2(y^2 + 6y) = x - 16\). Now, complete the square within the parentheses: take half of 6, square it to get 9, and add and subtract 9 inside the parentheses: \(2(y^2 + 6y + 9 - 9)\).
04
Write the Perfect Square
The expression \(y^2 + 6y + 9\) is a perfect square and can be written as \((y + 3)^2\). The equation is now \(2((y + 3)^2 - 9) = x - 16\), which simplifies to \(2(y + 3)^2 - 18 = x - 16\).
05
Simplify to Standard Form
Add 18 to both sides to isolate the perfect square: \(2(y + 3)^2 = x - 16 + 18\), which simplifies to \(2(y + 3)^2 = x + 2\). Finally, rearrange to match the standard form for horizontal parabolas: \((y + 3)^2 = \frac{1}{2}(x + 2)\).
06
Identify the Parabola Characteristics
In the equation \((y + 3)^2 = \frac{1}{2}(x + 2)\), the vertex \((h, k)\) is given by \((-2, -3)\), and the value \(4p = \frac{1}{2}\) indicates \(p = \frac{1}{8}\). The parabola opens to the right since \(x\) is isolated.
07
Graph the Parabola
Plot the vertex \((-2, -3)\) on the coordinate plane. As the parabola opens to the right, plot additional points by selecting \(y\) values around \(y = -3\), calculating corresponding \(x\) values using the standard form \((y + 3)^2 = \frac{1}{2}(x + 2)\). Connect these points smoothly to sketch the parabola.
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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 pivotal technique for rewriting quadratic equations into a more manageable form. In the context of parabolas, completing the square helps transform standard equations into a form that quickly reveals the vertex and the orientation of the parabola.
To complete the square for an expression like \(ay^2 + by\), focus first on factoring out any coefficient from the \(y^2\) term. Suppose we start with \(2(y^2 + 6y)\) — here, we factor 2 from the \(y^2\) and \(y\) terms, isolating the binomial \(y^2 + 6y\).
The next step is to transform \(y^2 + 6y\) into a perfect square trinomial. Take half of the coefficient 6 (from \(6y\)), which is 3, and then square it to get 9. Add and subtract 9 within the binomial to keep the equation balanced: \(y^2 + 6y + 9 - 9\).
This results in a neatly packaged perfect square: \((y + 3)^2\), which significantly aids in rewriting and analyzing the equation. Completing the square simplifies many mathematical tasks, particularly transforming equations to their vertex forms.
To complete the square for an expression like \(ay^2 + by\), focus first on factoring out any coefficient from the \(y^2\) term. Suppose we start with \(2(y^2 + 6y)\) — here, we factor 2 from the \(y^2\) and \(y\) terms, isolating the binomial \(y^2 + 6y\).
The next step is to transform \(y^2 + 6y\) into a perfect square trinomial. Take half of the coefficient 6 (from \(6y\)), which is 3, and then square it to get 9. Add and subtract 9 within the binomial to keep the equation balanced: \(y^2 + 6y + 9 - 9\).
This results in a neatly packaged perfect square: \((y + 3)^2\), which significantly aids in rewriting and analyzing the equation. Completing the square simplifies many mathematical tasks, particularly transforming equations to their vertex forms.
Vertex of a Parabola
The vertex plays a crucial role in understanding the position and direction of a parabola. When an equation is in vertex form, it highlights the vertex directly, making it easy to graph or analyze further.
In the context of the exercise, the transformed equation \((y + 3)^2 = \frac{1}{2}(x + 2)\) shows that the vertex \((h, k)\) is \((-2, -3)\). This can be identified directly from the terms \((y + 3)^2\) and \((x + 2)\).
The sign changes within the parentheses indicate their direction on the coordinate axes. Specifically, the \(y + 3\) term sets the vertical position, and the \(x + 2\) term sets the horizontal position of the vertex.
Recognizing the vertex becomes straightforward as it is always the point at which the parabola turns direction — in this case, from rising to falling or vice versa if it opens horizontally. This property helps in sketching a quick and proportionate graph of the parabola.
In the context of the exercise, the transformed equation \((y + 3)^2 = \frac{1}{2}(x + 2)\) shows that the vertex \((h, k)\) is \((-2, -3)\). This can be identified directly from the terms \((y + 3)^2\) and \((x + 2)\).
The sign changes within the parentheses indicate their direction on the coordinate axes. Specifically, the \(y + 3\) term sets the vertical position, and the \(x + 2\) term sets the horizontal position of the vertex.
Recognizing the vertex becomes straightforward as it is always the point at which the parabola turns direction — in this case, from rising to falling or vice versa if it opens horizontally. This property helps in sketching a quick and proportionate graph of the parabola.
Graphing Parabolas
Graphing a parabola involves several clear steps, which can seem daunting at first but become simpler with practice. Begin by finding the vertex, since this serves as the anchor point for the entire curve.
In the exercise, the given vertex is \((-2, -3)\). Start by plotting this on a coordinate plane. With the established vertex, determine the parabola's orientation and width from the standard form \((y + 3)^2 = \frac{1}{2}(x + 2)\).
Notice that the equation's form — where \(y\) is squared — indicates a horizontal opening direction. Moreover, the coefficient \(\frac{1}{2}\) associated with \(x\) in this equation implies that the parabola is wider compared to if the coefficient were smaller.
To graph accurately, choose several values of \(y\) around \(y = -3\). Calculate the corresponding \(x\) values to these \(y\) values using the equation \((y + 3)^2 = \frac{1}{2}(x + 2)\). Plot these points accurately and smoothly connect them to shape the full parabola. Watching the curve take shape provides clear insight into the function's behavior, making this process an interesting exploration of geometry.
In the exercise, the given vertex is \((-2, -3)\). Start by plotting this on a coordinate plane. With the established vertex, determine the parabola's orientation and width from the standard form \((y + 3)^2 = \frac{1}{2}(x + 2)\).
Notice that the equation's form — where \(y\) is squared — indicates a horizontal opening direction. Moreover, the coefficient \(\frac{1}{2}\) associated with \(x\) in this equation implies that the parabola is wider compared to if the coefficient were smaller.
To graph accurately, choose several values of \(y\) around \(y = -3\). Calculate the corresponding \(x\) values to these \(y\) values using the equation \((y + 3)^2 = \frac{1}{2}(x + 2)\). Plot these points accurately and smoothly connect them to shape the full parabola. Watching the curve take shape provides clear insight into the function's behavior, making this process an interesting exploration of geometry.
