Chapter 2: Problem 49
In Exercises \(45-54,\) find the vertex of the parabola associated with each quadratic function. $$f(x)=-\frac{2}{5} x^{2}+\frac{3}{7} x+2$$
Short Answer
Expert verified
The vertex is \( \left( \frac{15}{28}, \frac{829}{392} \right) \).
Step by step solution
01
Identify the standard form of a quadratic function
A quadratic function is typically given in the standard form: \( f(x) = ax^2 + bx + c \). Here, we have \( a = -\frac{2}{5} \), \( b = \frac{3}{7} \), and \( c = 2 \).
02
Understand the formula for the vertex of a parabola
The vertex of a parabola given by the quadratic function \( ax^2 + bx + c \) can be found using the formula: \( (h, k) \), where \( h = -\frac{b}{2a} \).
03
Calculate h, the x-coordinate of the vertex
Substitute the values of \( a \) and \( b \) into \( h = -\frac{b}{2a} \). This gives us:\[h = -\frac{\frac{3}{7}}{2(-\frac{2}{5})} = \frac{3}{7} \times \frac{5}{4} = \frac{15}{28}.\]
04
Calculate k, the y-coordinate of the vertex
Substitute \( h = \frac{15}{28} \) back into the original function to find \( k \):\[f\left(\frac{15}{28}\right) = -\frac{2}{5}\left(\frac{15}{28}\right)^2 + \frac{3}{7}\left(\frac{15}{28}\right) + 2.\]First, compute \( \left(\frac{15}{28}\right)^2 = \frac{225}{784} \).Then, evaluate each term:- \(-\frac{2}{5} \times \frac{225}{784} = -\frac{450}{3920} = -\frac{45}{392}.\)- \( \frac{3}{7} \times \frac{15}{28} = \frac{45}{196}. \)Combine the terms:\[k = -\frac{45}{392} + \frac{45}{196} + 2 = -\frac{45}{392} + \frac{90}{392} + \frac{784}{392} = \frac{829}{392}.\]
05
State the vertex
The vertex of the parabola is \( \left( \frac{15}{28}, \frac{829}{392} \right) \).
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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 characterized by its highest degree of 2. It takes the general form:
- \( f(x) = ax^2 + bx + c \)
- \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \)
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
- \( f(x) = -\frac{2}{5} x^2 + \frac{3}{7} x + 2 \)
Vertex Formula
The vertex of a quadratic function is a crucial point on its graph where it reaches its maximum or minimum value. This point is always on the parabola's axis of symmetry, dividing it into two mirror-image halves.
To calculate the vertex, we employ the vertex formula. For any quadratic function in the form \( ax^2 + bx + c \), the vertex \((h, k)\) has:
To calculate the vertex, we employ the vertex formula. For any quadratic function in the form \( ax^2 + bx + c \), the vertex \((h, k)\) has:
- \( h = -\frac{b}{2a} \)
- Using the given function \( f(x) = -\frac{2}{5}x^2 + \frac{3}{7}x + 2 \), we calculated \( h \) as \( \frac{15}{28} \).
- Then, plugging \( h \) back, we computed \( k \) to be \( \frac{829}{392} \).
Parabola
The parabola is the U-shaped graph that represents a quadratic function. Its orientation and width are influenced by the coefficient \( a \) in the quadratic equation:For upward-opening parabolas, the vertex is at the minimum point. For downward-opening parabolas, like in the given function \( f(x) = -\frac{2}{5} x^2 + \frac{3}{7} x + 2 \), the vertex is at the maximum point. The axis of symmetry is a vertical line passing through the vertex, dividing the parabola into two symmetrical halves and can be expressed with the equation \( x = h \), where \( h \) is the x-coordinate of the vertex.
Each parabola has a directrix, a horizontal line opposite from the vertex's opening and describes how 'wide' or 'narrow' the parabola is. Understanding these features of a parabola are key in graphing and analyzing quadratic functions.
- When \( |a| \) is large, the parabola is narrow.
- When \( |a| \) is small, the parabola is wide.
Each parabola has a directrix, a horizontal line opposite from the vertex's opening and describes how 'wide' or 'narrow' the parabola is. Understanding these features of a parabola are key in graphing and analyzing quadratic functions.