College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter4: Polynomial And Rational Functions
Section4.CT: Chapter Test
Problem 6CT: Assume that an object tossed vertically upward reaches a height of h feet after t seconds, where...
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![### Write an equation (any form) for the quadratic graphed below
#### Description of the Graph:
The graph displays a parabola that opens downwards. It is positioned on a coordinate plane with the x-axis ranging from -5 to 5 and the y-axis ranging from -5 to 5.
- The vertex of the parabola is at the point (-2, 4).
- The parabola intersects the x-axis at two points: \(x = -5\) and \(x = 1\).
- The y-axis passing through the y-coordinate 0 intersects the parabola at the point \(y = -3\), when \(x = 0\).
#### Diagram Explanation:
The graph plots a quadratic function on a coordinate plane. The highest point of the graph (vertex) is at (-2, 4). This vertex indicates that the function has a maximum value at this point since the graph opens downwards. The width and direction of the parabola are consistent with the standard shape of quadratic graphs.
#### Diagram:
[Here you would include an illustration of the graph similar to the one described above.]
#### Equation (Vertex Form):
Given the vertex form of a quadratic equation:
\[ y = a(x - h)^2 + k \]
Where:
- \( (h, k) \) is the vertex of the parabola
- \( a \) is the leading coefficient that determines the width and direction of the parabola
Since the vertex is at (-2, 4):
\[ y = a(x + 2)^2 + 4 \]
To find the value of \( a \), use another point from the graph. Let's use the point \( (1, 0) \):
\[ 0 = a(1 + 2)^2 + 4 \]
\[ 0 = a(3)^2 + 4 \]
\[ 0 = 9a + 4 \]
\[ -4 = 9a \]
\[ a = -\frac{4}{9} \]
Therefore, the equation in vertex form is:
\[ y = -\frac{4}{9}(x + 2)^2 + 4 \]
#### Question Help:
To further understand how to derive the equation of the parabola, you can watch the following instructional videos for additional guidance:
- Video 1 [Link to video]
- Video 2 [Link to video]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5253470b-a88a-44a2-95c4-a44ae3e11575%2F5a3b77ac-2926-43e0-a57e-614b4e06ddcc%2Fs7hnod_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Write an equation (any form) for the quadratic graphed below
#### Description of the Graph:
The graph displays a parabola that opens downwards. It is positioned on a coordinate plane with the x-axis ranging from -5 to 5 and the y-axis ranging from -5 to 5.
- The vertex of the parabola is at the point (-2, 4).
- The parabola intersects the x-axis at two points: \(x = -5\) and \(x = 1\).
- The y-axis passing through the y-coordinate 0 intersects the parabola at the point \(y = -3\), when \(x = 0\).
#### Diagram Explanation:
The graph plots a quadratic function on a coordinate plane. The highest point of the graph (vertex) is at (-2, 4). This vertex indicates that the function has a maximum value at this point since the graph opens downwards. The width and direction of the parabola are consistent with the standard shape of quadratic graphs.
#### Diagram:
[Here you would include an illustration of the graph similar to the one described above.]
#### Equation (Vertex Form):
Given the vertex form of a quadratic equation:
\[ y = a(x - h)^2 + k \]
Where:
- \( (h, k) \) is the vertex of the parabola
- \( a \) is the leading coefficient that determines the width and direction of the parabola
Since the vertex is at (-2, 4):
\[ y = a(x + 2)^2 + 4 \]
To find the value of \( a \), use another point from the graph. Let's use the point \( (1, 0) \):
\[ 0 = a(1 + 2)^2 + 4 \]
\[ 0 = a(3)^2 + 4 \]
\[ 0 = 9a + 4 \]
\[ -4 = 9a \]
\[ a = -\frac{4}{9} \]
Therefore, the equation in vertex form is:
\[ y = -\frac{4}{9}(x + 2)^2 + 4 \]
#### Question Help:
To further understand how to derive the equation of the parabola, you can watch the following instructional videos for additional guidance:
- Video 1 [Link to video]
- Video 2 [Link to video]
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