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Chapter 9: Problem 34

Graph each parabola. Plot at least two points as well as the vertex. Give the vertex, axis, domain, and range . $$ f(x)=-\frac{2}{3}(x+2)^{2}+1 $$

Short Answer

Expert verified
Vertex: (-2, 1), Axis: x = -2, Domain: \(\mathbb{R}\), Range: \((-\infty, 1 ]\).

Step by step solution

01

Identify the vertex

The vertex formula for a parabola in the form \(f(x) = a(x-h)^2 + k\) is given directly by \((h, k)\). Here, \(a = -\frac{2}{3}\), \(h = -2\), and \(k = 1\). Thus, the vertex is at \((-2, 1)\).
02

Determine the axis of symmetry

The axis of symmetry for a parabola given in the form \(f(x) = a(x-h)^2 + k\) is a vertical line that passes through the vertex. For this parabola, the axis of symmetry is \(x = -2\).
03

Find additional points

To graph the parabola, another key point is needed. Choose \(x = -1\) and solve for \(f(x)\):\[f(-1) = -\frac{2}{3}(-1 + 2)^2 + 1 = -\frac{2}{3}(1)^2 + 1 = -\frac{2}{3} + 1 = \frac{1}{3}\]. So, one point is \((-1, \frac{1}{3})\). Choosing another value, \(x = -3\) gives: \[f(-3) = -\frac{2}{3}(-3 + 2)^2 + 1 = -\frac{2}{3}(-1)^2 + 1 = -\frac{2}{3} + 1 = \frac{1}{3}\]. Another point is \((-3, \frac{1}{3})\).
04

State the domain and range

The domain of all parabolas is all real numbers: \(\mathbb{R}\). The range of this downward-opening parabola is all y-values less than or equal to 1, because the maximum value of \(f(x)\) (the value at the vertex) is 1. Thus, the range is \(( -\infty, 1 ]\).
05

Plot and graph

Plot the vertex \((-2, 1)\), points \((-1, \frac{1}{3})\) and \((-3, \frac{1}{3})\) on a coordinate plane. Draw a smooth curve through these points to represent the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertex of a parabola
The vertex of a parabola is a crucial point that represents the parabola's highest or lowest point, depending on its direction. For any parabola written in the form \(f(x) = a(x-h)^2 + k\), the vertex can be easily identified as the point \((h, k)\). In this exercise, we have \(a = -\frac{2}{3}\), \(h = -2\), and \(k = 1\). Therefore, the vertex is at \((-2, 1)\). This point marks the highest point of the parabola because the parabola opens downwards (indicated by the negative \(a\) value).
axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. This line always passes through the vertex. For parabolas of the form \(f(x) = a(x-h)^2 + k\), the axis of symmetry is given by the equation \(x = h\). In our case, the axis of symmetry is the line \(x = -2\). This vertical line helps us understand how the parabola is balanced and aids in plotting points on either side of it.
domain and range
The domain of a parabola includes all possible x-values that can be input into the function. For any quadratic function, this is all real numbers, represented as \(\backslashmathbb{R}\). However, the range is the set of all y-values that the parabola can output. Since our parabola opens downwards with a vertex at \(( -2, 1 )\), the highest y-value is 1. Therefore, the range includes all y-values less than or equal to 1, written as \( ( -\infty, 1 ]\). This tells us how far the parabola stretches vertically.
plotting points
To graph a parabola accurately, it's essential to plot points around the vertex. Start by plotting the vertex, such as \((-2, 1) \). Then, choose additional x-values to find corresponding y-values. For instance, when \(x = -1\), we found \(f(-1) = \frac{1}{3}\), giving the point \((-1, \frac{1}{3})\). Similarly, for \(x = -3\), we get another point \((-3, \frac{1}{3}) \). By plotting these points and others if needed, and drawing a smooth curve through them, we can visualize the parabola's shape. This step helps to ensure that the graph is accurate and complete.

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Most popular questions from this chapter

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