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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 7 - Conic SectionsProblema 67
Capitolo 8, Problema 67

Graph each semiellipse. y = -√16 - 4x²

Guida verificata passo dopo passo
1
Recognize that the given equation is \(y = \sqrt{16 - 4x^{2}}\), which represents the upper half of an ellipse because \(y\) is defined as the positive square root.
Rewrite the equation by squaring both sides to eliminate the square root: \(y^{2} = 16 - 4x^{2}\).
Rearrange the equation to standard ellipse form by moving all terms to one side: \(4x^{2} + y^{2} = 16\).
Divide the entire equation by 16 to normalize it: \(\frac{4x^{2}}{16} + \frac{y^{2}}{16} = 1\), which simplifies to \(\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1\).
Identify the ellipse parameters: the semi-major axis length \(a = 4\) (along the y-axis) and the semi-minor axis length \(b = 2\) (along the x-axis). Since \(y = \sqrt{16 - 4x^{2}}\) only gives the upper half, graph the ellipse only for \(y \geq 0\).

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Equation of an Ellipse

An ellipse is a set of points where the sum of distances to two foci is constant. Its standard form is (x²/a²) + (y²/b²) = 1. The given equation y = √(16 - 4x²) represents the upper half of an ellipse after rearranging to fit this form.
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Foci and Vertices of an Ellipse

Graphing Functions with Square Roots

When graphing y = √(expression), the output y is always non-negative, representing only the upper part of the curve. This means the graph is a semiellipse (half ellipse) above the x-axis, and the domain is restricted to values where the expression inside the root is non-negative.
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Imaginary Roots with the Square Root Property

Domain and Range of the Function

The domain consists of all x-values for which the expression under the square root is ≥ 0. For y = √(16 - 4x²), this means 16 - 4x² ≥ 0, limiting x between -2 and 2. The range is y ≥ 0 since the square root function outputs non-negative values.
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4:22
Domain & Range of Transformed Functions
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