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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 7 - Conic SectionsProblem 61
Chapter 8, Problem 61

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?
x=−4(y−1)2+3x = - 4(y - 1)^2 + 3

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Identify the given equation: x = -4(y - 1)2 + 3. This is a parabola expressed in terms of y.
Recognize the vertex form of the parabola. Here, the vertex is at (3, 1) because the equation is in the form x = a(y - k)^2 + h, where (h, k) is the vertex.
Determine the direction the parabola opens by looking at the coefficient of the squared term, a = -4. Since a is negative, the parabola opens to the left (towards decreasing x values).
Find the domain by considering the range of x values. Since the parabola opens left from the vertex at x = 3, the domain is all x values less than or equal to 3, or x \(\leq\) 3.
Find the range by considering all possible y values. Because the parabola is symmetric about y = 1 and opens horizontally, y can take any real value, so the range is (-\(\infty\), \(\infty\)). Finally, determine if the relation is a function: since for some x values there are multiple y values, it is not a function.

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Key Concepts

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

Vertex Form of a Parabola

The vertex form of a parabola expresses the equation in a way that reveals its vertex, the highest or lowest point. For example, x = a(y - k)^2 + h shows a parabola with vertex at (h, k). Understanding the vertex helps identify key features like the parabola's position and symmetry.
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Vertex Form

Direction of Opening of a Parabola

The sign and variable squared in the equation determine the parabola's direction. If the squared term is y, the parabola opens horizontally; if x, it opens vertically. The coefficient's sign indicates whether it opens left/right or up/down, which affects the domain and range.
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Horizontal Parabolas

Domain and Range of Relations and Functions

The domain is the set of all possible input values (x or y), and the range is the set of all possible output values. For parabolas, these depend on the vertex and opening direction. Determining if the relation is a function involves checking if each input corresponds to exactly one output.
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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\(\end{cases}\)

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Textbook Question

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

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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.

{x2+y2=1x2+9y2=9\(\begin{cases}\)x^2 + y^2 = 1 \(\x\)^2 + 9y^2 = 9\(\end{cases}\)

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Textbook Question

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

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Textbook Question

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?

y=−x2+4x−3y=-x^2+4x-3


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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.

{(y−2)2 =x+4y=−(12)x\(\left\)\{\(\begin{array}{l}\]\left\)(y-2\(\right\))^2\(\text{ }\)=x+4\\ y=-\(\text{(}\[\frac\)12\(\text{)}\)x\(\end{array}\]\right\).

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