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Magazine Chapter 2: Problem 48
In Exercises \(45-54,\) find the vertex of the parabola associated with each quadratic function. $$f(x)=-\frac{1}{2} x^{2}+\frac{2}{3} x+4$$
Short Answer
Expert verified
The vertex of the parabola is \(\left(\frac{2}{3}, \frac{38}{9}\right)\).
Step by step solution
01
Identify Coefficients
To find the vertex of the quadratic function, we first need to identify the coefficients from the quadratic formula \( f(x) = ax^2 + bx + c \). In this case, \( a = -\frac{1}{2}, b = \frac{2}{3}, \text{ and } c = 4 \).
02
Find the x-coordinate of the Vertex
The x-coordinate of the vertex of a parabola given by \( f(x) = ax^2 + bx + c \) is found using the formula \( x = \frac{-b}{2a} \). Substituting the given values, we calculate:\[x = \frac{-\frac{2}{3}}{2(-\frac{1}{2})} = \frac{-\frac{2}{3}}{-1} = \frac{2}{3}\]
03
Find the y-coordinate of the Vertex
To find the y-coordinate of the vertex, substitute the x-coordinate back into the original function. Substitute \( x = \frac{2}{3} \) into \( f(x) \):\[f\left(\frac{2}{3}\right) = -\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{2}{3}\left(\frac{2}{3}\right) + 4\]Calculate each term:- \(-\frac{1}{2}\left(\frac{4}{9}\right) = -\frac{2}{9} \)- \( \frac{4}{9} \)Adding these with \( 4 \), we find the y-coordinate:\[f\left(\frac{2}{3}\right) = -\frac{2}{9} + \frac{4}{9} + 4 = \frac{2}{9} + 4 = \frac{2}{9} + \frac{36}{9} = \frac{38}{9}\]
04
Write the Vertex
The vertex of the parabola, written as \((h, k)\), is represented by the coordinates \( x \) and \( f(x) \) from the previous steps. With \( h = \frac{2}{3} \) and \( k = \frac{38}{9} \), the vertex of the parabola is:\[\left(\frac{2}{3}, \frac{38}{9}\right)\]
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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, specifically of degree 2. It is typically expressed in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
Quadratic functions create a U-shaped curve known as a parabola when graphed on a coordinate plane. This curve can open either upwards or downwards:
Quadratic functions create a U-shaped curve known as a parabola when graphed on a coordinate plane. This curve can open either upwards or downwards:
- If the coefficient \( a \) is positive, the parabola opens upwards.
- If \( a \) is negative, it opens downwards.
Vertex Form
Quadratic functions are often transformed into a vertex form to easily identify characteristics of the parabola. The vertex form of a quadratic function is:\[ f(x) = a(x-h)^2 + k \]In this form, \( h \) and \( k \) represent the coordinates of the vertex, \((h, k)\). This can be particularly useful because it visually and algebraically highlights the vertex.
- In the expression, \( (x-h) \) is horizontally shifting the parabola, effectively moving the vertex along the x-axis.
- The \( +k \) shifts the parabola vertically, adjusting the vertex along the y-axis.
Coefficients Identification
Identifying coefficients of a quadratic function is the first step in understanding its graph. In the standard form \( f(x) = ax^2 + bx + c \), each coefficient plays a different role:
- \( a \) affects the width and direction of the parabola.
- \( b \) influences the parabola's position along the x-axis.
- \( c \) indicates where the parabola crosses the y-axis (initial value).
X-Coordinate Calculation
To find the x-coordinate of the vertex of a parabola in standard form, use the formula:\[ x = \frac{-b}{2a} \]This formula efficiently finds the vertex鈥檚 x-coordinate because the vertex lies on the parabola's axis of symmetry.
For the function \( f(x) = -\frac{1}{2}x^2 + \frac{2}{3}x + 4 \), substituting the values gives:\[ x = \frac{-\frac{2}{3}}{2(-\frac{1}{2})} \]As calculated:\[ x = \frac{-\frac{2}{3}}{-1} = \frac{2}{3} \]This simplification highlights where the vertex lands on the x-axis, which is essential for accurately plotting the parabola.
For the function \( f(x) = -\frac{1}{2}x^2 + \frac{2}{3}x + 4 \), substituting the values gives:\[ x = \frac{-\frac{2}{3}}{2(-\frac{1}{2})} \]As calculated:\[ x = \frac{-\frac{2}{3}}{-1} = \frac{2}{3} \]This simplification highlights where the vertex lands on the x-axis, which is essential for accurately plotting the parabola.
Y-Coordinate Calculation
After determining the x-coordinate, calculate the y-coordinate of the vertex by substituting back into the quadratic function:\[ f\left(\frac{2}{3}\right) = -\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{2}{3}\left(\frac{2}{3}\right) + 4 \]Handle each term carefully:
- The squared term contributes to the downward opening because of the negative \( a \).
- The linear term shifts the function along the y-axis.
