Chapter 2: Problem 19
In Exercises \(9-22,\) graph the quadratic function, which is given in standard form. $$f(x)=\left(x-\frac{1}{3}\right)^{2}+\frac{1}{9}$$
Short Answer
Expert verified
The parabola has a vertex at \((\frac{1}{3}, \frac{1}{9})\) and opens upwards.
Step by step solution
01
Recognize the form of the quadratic function
The function is given as \(f(x) = \left(x - \frac{1}{3}\right)^2 + \frac{1}{9}\). This is in the form \(f(x) = (x - h)^2 + k\), where \(h\) and \(k\) are real numbers.
02
Identify the vertex
From the equation \(f(x) = \left(x - \frac{1}{3}\right)^2 + \frac{1}{9}\), it is clear that the vertex of the parabola is \((h, k) = (\frac{1}{3}, \frac{1}{9})\).
03
Determine the direction of the parabola
The quadratic function \(f(x) = \left(x - \frac{1}{3}\right)^2 + \frac{1}{9}\) has a positive coefficient in front of \((x-h)^2\), which is 1. This means the parabola opens upwards.
04
Find additional points for graphing
Choose a few x-values around the vertex, such as \(x = 0\) and \(x = 1\), to find additional points. Calculate:- \(f(0) = \left(0 - \frac{1}{3}\right)^2 + \frac{1}{9} = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}\)- \(f(1) = \left(1 - \frac{1}{3}\right)^2 + \frac{1}{9} = \left(\frac{2}{3}\right)^2 + \frac{1}{9} = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}\)
05
Plot the graph
Using the vertex \((\frac{1}{3}, \frac{1}{9})\) and the other points calculated \((0, \frac{2}{9})\) and \((1, \frac{5}{9})\), plot these points on a graph. Since the parabola opens upwards and is symmetric around the vertex, you can also plot the point \((\frac{2}{3}, \frac{2}{9})\) as a reflection of \((0, \frac{2}{9})\). Draw a smooth curve 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 Form
Understanding the vertex form of a quadratic function is crucial for graphing. The vertex form is given by the equation \( f(x) = a(x - h)^2 + k \). This equation makes it easy to identify the vertex of the parabola, which is the point \((h, k)\). In our exercise, the function \( f(x) = (x - \frac{1}{3})^2 + \frac{1}{9} \) follows this form. Here, the vertex \((h, k)\) is at \( (\frac{1}{3}, \frac{1}{9})\). The vertex is the lowest point of the parabola if it opens upwards, or the highest if it opens downwards.
- Vertex: Point \((h, k)\) is crucial for graph symmetry.
- Helps in determining the minimum or maximum value of the function.
Parabola Direction
The direction of the parabola is determined by the coefficient \( a \) in the vertex form equation \( f(x) = a(x - h)^2 + k \). If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards. For the function \( f(x) = (x - \frac{1}{3})^2 + \frac{1}{9} \), the coefficient \( a \) is 1, which means the parabola opens upwards.
- An upward-opening parabola has a minimum point at the vertex.
- A downward-opening parabola has a maximum point at the vertex.
Plotting Points
To accurately graph a quadratic function, plotting additional points around the vertex is essential. These points help define the shape and location of the parabola on a graph. In this case, we've calculated several points:
- For \( x = 0 \), \( f(0) = \frac{2}{9} \)
- For \( x = 1 \), \( f(1) = \frac{5}{9} \)