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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 59
Chapter 8, Problem 59

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

Verified step by step guidance
1
Identify the given quadratic function: y = -x2 + 4x - 3.
Determine the direction in which the parabola opens by looking at the coefficient of x2. Since it is negative (-1), the parabola opens downward.
Find the vertex of the parabola using the vertex formula for x: x = -\(\frac{b}{2a}\), where a = -1 and b = 4. Calculate x = -\(\frac{4}{2(-1)}\).
Substitute the x-value of the vertex back into the original equation to find the y-coordinate of the vertex, which gives the maximum value of y because the parabola opens downward.
Determine the domain and range: The domain of any quadratic function is all real numbers, so \(\text{Domain}\) = (-\(\infty\), \(\infty\)). The range is all y-values less than or equal to the vertex's y-coordinate, so \(\text{Range}\) = (-\(\infty\), y_{vertex}]. Since each x corresponds to exactly one y, the relation is a function.

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Key Concepts

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

Vertex of a Parabola

The vertex is the highest or lowest point on a parabola, found using the formula x = -b/(2a) for a quadratic y = ax^2 + bx + c. It helps identify the maximum or minimum value of the function, which is crucial for determining the range.
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Horizontal Parabolas

Direction of the Parabola

The direction a parabola opens depends on the coefficient 'a' in y = ax^2 + bx + c. If 'a' is positive, it opens upward; if negative, downward. This direction indicates whether the vertex is a maximum or minimum point, affecting the range of the relation.
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Horizontal Parabolas

Domain and Range of Quadratic Functions

The domain of a quadratic function is all real numbers since x can take any value. The range depends on the vertex and the parabola's direction: if it opens downward, the range is all y-values less than or equal to the vertex's y-coordinate; if upward, all y-values greater than or equal to it.
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Domain & Range of Transformed Functions
Related Practice
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?

x=−4(y−1)2+3x = - 4(y - 1)^2 + 3


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. 25x²+4y² – 150x + 32y + 189 = 0

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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

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\(\end{cases}\)

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 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\).