/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Graph each function. $$y=-\fra... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 16

Graph each function. $$y=-\frac{1}{3} x^{2}$$

Short Answer

Expert verified
The graph is a downward-opening parabola with vertex at (0,0) and additional points at (-3,-3) and (3,-3).

Step by step solution

01

- Identify the basic form of the quadratic function

The given function is a quadratic function of the form \(y = ax^2\). In this case, \(a = -\frac{1}{3}\).
02

- Determine the direction of the parabola

Since \(a = -\frac{1}{3}\) is negative, the parabola opens downward.
03

- Find the vertex of the parabola

For a function in the form \(y = ax^2\), the vertex is at the origin, \( (0,0) \).
04

- Plot the vertex

Plot the point (0,0) on the coordinate plane.
05

- Choose additional points to plot the parabola

Select x-values to substitute into the function and find corresponding y-values. For example, \( x = -3 \) and \( x = 3 \).
06

- Calculate and plot additional points

For \( x = -3\): \(-\frac{1}{3}(-3)^2 = -\frac{1}{3} \cdot 9 = -3\). So, the point is (-3,-3). For \( x = 3\): \(-\frac{1}{3}(3)^2 = -\frac{1}{3} \cdot 9 = -3\). So, the point is (3,-3). Plot these points.
07

- Sketch the parabola

Using the points (0,0), (-3,-3), and (3,-3), draw a smooth curve to form the downward-opening parabola.

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Key Concepts

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

Quadratic Function
A quadratic function is a type of polynomial function that can be written in the standard form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The graph of a quadratic function is always a parabola. In the equation provided in the exercise, the function is given as \(y = - \frac{1}{3} x^2 \).
  • The term \(a\) determines the direction and the width of the parabola.
  • Since \(a = - \frac{1}{3}\) in the given function, the parabola opens downward (because \(a\) is negative).
  • The smaller the absolute value of \(a\), the wider the parabola will be. Here \(a = - \frac{1}{3}\), making the parabola relatively wide.
Parabola
A parabola is the U-shaped graph formed by plotting a quadratic function. Its key features include the direction it opens, which is determined by the sign of \(a\) in the quadratic function.The general characteristics of a parabola are:
  • If \(a > 0\), it opens upwards, resembling a 'smiling' shape.
  • If \(a < 0\), it opens downwards, resembling a 'frowning' shape.
In the provided exercise, since \(a = - \frac{1}{3}\), the parabola opens downward. To graph a parabola, you need to plot the vertex first and then plot additional points by selecting x-values, substituting them into the function to calculate corresponding y-values, and then drawing a smooth curve through these points.
Vertex
The vertex of a parabola is its highest or lowest point, depending on the direction the parabola opens. For the function \(y = ax^2\), the vertex is always at the origin, which is (0,0).
Here, the vertex is also (0,0). To find and plot the vertex:
  • Identify the value of \(a\) in the function.
  • If the equation is in the form \(y = ax^2\), like in this example, the vertex is at (0,0).
After plotting the vertex, we can choose additional points to define the shape of the parabola. By using x-values (eg. -3, 3), and substituting them back into the function, we get further points on the graph, such as (-3,-3) and (3,-3), which help shape the accurate curve.

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