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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 8, Problem 67

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

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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Domain & Range of Transformed Functions
Related Practice
Textbook Question

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.

{x225+y29=1y=3\(\begin{cases}\]\frac{x^2}{25}\) + \(\frac{y^2}{9}\) = 1 \(\y\) = 3\(\end{cases}\)

Textbook Question

The equation of the red ellipse in the figure shown is x^2/25 + y^2/9 =1Write the equation for each circle shown in the figure.

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Textbook Question

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. {x=(y+2)21(x2)2+(y+2)2=1\(\begin{cases}\)x = (y + 2)^2 - 1 \\(x - 2)^2 + (y + 2)^2 = 1\(\end{cases}\)

Textbook Question

Find the standard form of the equation of an ellipse with vertices at (0, -6) and (0, 6), passing through (2, 4).

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Textbook Question

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.

{x=y23x=y23y\(\left\)\{\(\begin{array}{l}\)x=y^2-3\\ x=y^2-3y\(\end{array}\]\right\).

Textbook Question

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.

{4x2+y2=42xy=2\(\begin{cases}\)4x^2 + y^2 = 4 \\2x - y = 2\(\end{cases}\)