교과서 질문
In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function?
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Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
8장, 문제 59
Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 36x2 +9y2 - 216x = 0
검증된 단계별 안내1
Start with the given equation: \(36x^{2} + 9y^{2} - 216x = 0\).
Group the \(x\) terms and the \(y\) terms together: \(36x^{2} - 216x + 9y^{2} = 0\).
Factor out the coefficients of the squared terms to make completing the square easier: \(36(x^{2} - 6x) + 9y^{2} = 0\).
Complete the square for the \(x\) terms inside the parentheses. Take half of the coefficient of \(x\) (which is \(-6\)), square it, and add inside the parentheses, remembering to balance the equation by adding the equivalent value outside: half of \(-6\) is \(-3\), and \((-3)^{2} = 9\). So add and subtract \(9\) inside the parentheses: \(36(x^{2} - 6x + 9 - 9) + 9y^{2} = 0\).
Rewrite the completed square and simplify the constants: \(36(x - 3)^{2} - 36 imes 9 + 9y^{2} = 0\). Then move the constant term to the other side and divide the entire equation by the constant to get the ellipse in standard form.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
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주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Completing the Square
Completing the square is a method used to rewrite quadratic expressions in the form (x - h)² or (y - k)² by adding and subtracting appropriate constants. This technique helps transform the given equation into a recognizable conic section form, such as the standard form of an ellipse.
추천 영상:
06:24
Solving Quadratic Equations by Completing the Square
Standard Form of an Ellipse
The standard form of an ellipse equation is (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center, and a and b are the lengths of the semi-major and semi-minor axes. Converting to this form allows easy identification of the ellipse's size, shape, and position.
추천 영상:
5:12Graph Ellipses at Origin
Foci of an Ellipse
The foci are two fixed points inside the ellipse such that the sum of distances from any point on the ellipse to the foci is constant. Their locations depend on the values of a, b, and the center, and are found using c² = |a² - b²|, where c is the distance from the center to each focus.
추천 영상:
5:30Foci and Vertices of an Ellipse
관련 실천
교과서 질문
Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 25x²+4y² – 150x + 32y + 189 = 0
827
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교과서 질문
Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? y2 + 6y - x + 5 = 0
892
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교과서 질문
Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
812
views
교과서 질문
In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function?
732
views
교과서 질문
In Exercises 63–68, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
650
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