Chapter 2: Problem 45
In Exercises \(45-54,\) find the vertex of the parabola associated with each quadratic function. $$f(x)=33 x^{2}-2 x+15$$
Short Answer
Expert verified
The vertex of the parabola is \( \left( \frac{1}{33}, 14.9697 \right) \).
Step by step solution
01
Identify the Coefficients
The quadratic function is given by the equation \( f(x) = 33x^2 - 2x + 15 \). Let's identify the coefficients: \( a = 33 \), \( b = -2 \), and \( c = 15 \). These will be used in the formula for the vertex of a parabola.
02
Use the Vertex Formula for the x-coordinate
The x-coordinate of the vertex for a parabola given by \( f(x) = ax^2 + bx + c \) is calculated using \( x = -\frac{b}{2a} \). Substitute the values of \( a \) and \( b \) into this formula: \[ x = -\frac{-2}{2 \times 33} = \frac{2}{66} = \frac{1}{33} \].
03
Calculate the y-coordinate
Substitute \( x = \frac{1}{33} \) back into the original function \( f(x) \) to find the y-coordinate of the vertex. \[ f\left(\frac{1}{33}\right) = 33 \left(\frac{1}{33}\right)^2 - 2 \left(\frac{1}{33}\right) + 15 \]. Simplify this to get: \[ f\left(\frac{1}{33}\right) = \frac{33}{1089} - \frac{2}{33} + 15 \]. Further simplify to find \( f\left(\frac{1}{33}\right) \).
04
Simplify the Expression
Simplify \( \frac{33}{1089} \) and \( \frac{2}{33} \): \( \frac{33}{1089} = \frac{1}{33} \). So, \[ f\left(\frac{1}{33}\right) = \frac{1}{33} - \frac{2}{33} + 15 = -\frac{1}{33} + 15 \]. The y-coordinate becomes: \[ f\left(\frac{1}{33}\right) = 15 - \frac{1}{33} \].
05
Write the Vertex Coordinates
After simplifying, the coordinates of the vertex are \( \left( \frac{1}{33}, 14.9697 \right) \), where \( 14.9697 \) is the approximate value of \( 15 - \frac{1}{33} \). Therefore, the vertex of the parabola is \( \left( \frac{1}{33}, 14.9697 \right) \).
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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 is represented by the equation \( f(x) = ax^2 + bx + c \). This equation characterizes a unique curve on a graph known as a parabola. Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \), which ensures the function is indeed quadratic.
- The term \( ax^2 \) is the quadratic term and determines the parabola's width and the direction it opens.
- The term \( bx \) is the linear term influencing the slope of the parabola.
- Constant \( c \) is the y-intercept of the graph, showing where the parabola crosses the y-axis.
Vertex Formula
The vertex of a parabola is its highest or lowest point, depending on the direction in which the parabola opens. For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex provides critical information about the graph. To find the vertex, we rely on the vertex formula:\[ x = -\frac{b}{2a} \]
- This equation gives the x-coordinate of the vertex.
- Once we have the x-coordinate, we substitute it back into the function to find the y-coordinate.
- The x-coordinate as \( x = -\frac{-2}{2 \times 33} = \frac{1}{33} \).
- To find the y-coordinate, substitute \( x = \frac{1}{33} \) into the function to get \( y \approx 14.9697 \).
Parabola
A parabola is the U-shaped graphical representation of a quadratic function. It exhibits symmetry, which is one of its defining characteristics. Whether the parabola opens upwards or downwards depends on the coefficient \( a \) in the quadratic function:
- If \( a > 0 \), the parabola opens upwards and the vertex is the minimum point.
- If \( a < 0 \), it opens downwards and the vertex is the maximum point.