Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To graph the equation [tex]\( h = -16t^2 + 80 \)[/tex], we can follow a series of steps:
### 1. Understand the Equation
The given equation [tex]\( h = -16t^2 + 80 \)[/tex] is a quadratic equation in the form [tex]\( h = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 80 \)[/tex]
This represents a parabola that opens downward because the coefficient of [tex]\( t^2 \)[/tex] (which is [tex]\( -16 \)[/tex]) is negative.
### 2. Determine Key Features of the Parabola
- Vertex: The vertex of a parabola defined by [tex]\( at^2 + bt + c \)[/tex] is at [tex]\( t = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 0 \)[/tex], so:
[tex]\[ t = -\frac{0}{2 \times -16} = 0 \][/tex]
Substituting [tex]\( t = 0 \)[/tex] back into the equation to find [tex]\( h \)[/tex]:
[tex]\[ h = -16(0)^2 + 80 = 80 \][/tex]
Therefore, the vertex is at [tex]\( (0, 80) \)[/tex].
- Axis of Symmetry: The axis of symmetry is a vertical line that goes through the vertex, which is [tex]\( t = 0 \)[/tex].
- Intercepts:
- Y-intercept: Set [tex]\( t = 0 \)[/tex]:
[tex]\[ h = -16(0)^2 + 80 = 80 \][/tex]
Thus, the y-intercept is [tex]\( (0, 80) \)[/tex].
- X-intercepts: Set [tex]\( h = 0 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 80 \][/tex]
Rearranging:
[tex]\[ 16t^2 = 80 \][/tex]
[tex]\[ t^2 = \frac{80}{16} = 5 \][/tex]
[tex]\[ t = \pm \sqrt{5} \][/tex]
Therefore, the x-intercepts are [tex]\( (\sqrt{5}, 0) \)[/tex] and [tex]\( (-\sqrt{5}, 0) \)[/tex].
### 3. Plot Points and Draw the Graph
To graph the quadratic function, we will plot a few key points and then sketch the curve:
- Vertex: [tex]\( (0, 80) \)[/tex]
- X-intercepts: [tex]\( (\sqrt{5}, 0) \approx (2.24, 0) \)[/tex] and [tex]\( (-\sqrt{5}, 0) \approx (-2.24, 0) \)[/tex]
- Additional points for better accuracy:
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ h = -16(1)^2 + 80 = -16 + 80 = 64 \][/tex]
Hence, point [tex]\( (1, 64) \)[/tex].
- At [tex]\( t = -1 \)[/tex]:
[tex]\[ h = -16(-1)^2 + 80 = -16 + 80 = 64 \][/tex]
Hence, point [tex]\( (-1, 64) \)[/tex].
- More points can be calculated similarly if needed to show the curve properly.
### 4. Sketch the Parabola
Using the calculated points, we can now sketch the graph of [tex]\( h = -16t^2 + 80 \)[/tex]:
1. Plot the vertex [tex]\( (0, 80) \)[/tex].
2. Plot the x-intercepts [tex]\( (2.24, 0) \)[/tex] and [tex]\( (-2.24, 0) \)[/tex].
3. Plot additional points [tex]\( (1, 64) \)[/tex] and [tex]\( (-1, 64) \)[/tex].
4. Draw a smooth curve through these points, ensuring the parabola opens downward.
This provides a clear visual representation of the equation [tex]\( h = -16t^2 + 80 \)[/tex].
### 1. Understand the Equation
The given equation [tex]\( h = -16t^2 + 80 \)[/tex] is a quadratic equation in the form [tex]\( h = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 80 \)[/tex]
This represents a parabola that opens downward because the coefficient of [tex]\( t^2 \)[/tex] (which is [tex]\( -16 \)[/tex]) is negative.
### 2. Determine Key Features of the Parabola
- Vertex: The vertex of a parabola defined by [tex]\( at^2 + bt + c \)[/tex] is at [tex]\( t = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 0 \)[/tex], so:
[tex]\[ t = -\frac{0}{2 \times -16} = 0 \][/tex]
Substituting [tex]\( t = 0 \)[/tex] back into the equation to find [tex]\( h \)[/tex]:
[tex]\[ h = -16(0)^2 + 80 = 80 \][/tex]
Therefore, the vertex is at [tex]\( (0, 80) \)[/tex].
- Axis of Symmetry: The axis of symmetry is a vertical line that goes through the vertex, which is [tex]\( t = 0 \)[/tex].
- Intercepts:
- Y-intercept: Set [tex]\( t = 0 \)[/tex]:
[tex]\[ h = -16(0)^2 + 80 = 80 \][/tex]
Thus, the y-intercept is [tex]\( (0, 80) \)[/tex].
- X-intercepts: Set [tex]\( h = 0 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 80 \][/tex]
Rearranging:
[tex]\[ 16t^2 = 80 \][/tex]
[tex]\[ t^2 = \frac{80}{16} = 5 \][/tex]
[tex]\[ t = \pm \sqrt{5} \][/tex]
Therefore, the x-intercepts are [tex]\( (\sqrt{5}, 0) \)[/tex] and [tex]\( (-\sqrt{5}, 0) \)[/tex].
### 3. Plot Points and Draw the Graph
To graph the quadratic function, we will plot a few key points and then sketch the curve:
- Vertex: [tex]\( (0, 80) \)[/tex]
- X-intercepts: [tex]\( (\sqrt{5}, 0) \approx (2.24, 0) \)[/tex] and [tex]\( (-\sqrt{5}, 0) \approx (-2.24, 0) \)[/tex]
- Additional points for better accuracy:
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ h = -16(1)^2 + 80 = -16 + 80 = 64 \][/tex]
Hence, point [tex]\( (1, 64) \)[/tex].
- At [tex]\( t = -1 \)[/tex]:
[tex]\[ h = -16(-1)^2 + 80 = -16 + 80 = 64 \][/tex]
Hence, point [tex]\( (-1, 64) \)[/tex].
- More points can be calculated similarly if needed to show the curve properly.
### 4. Sketch the Parabola
Using the calculated points, we can now sketch the graph of [tex]\( h = -16t^2 + 80 \)[/tex]:
1. Plot the vertex [tex]\( (0, 80) \)[/tex].
2. Plot the x-intercepts [tex]\( (2.24, 0) \)[/tex] and [tex]\( (-2.24, 0) \)[/tex].
3. Plot additional points [tex]\( (1, 64) \)[/tex] and [tex]\( (-1, 64) \)[/tex].
4. Draw a smooth curve through these points, ensuring the parabola opens downward.
This provides a clear visual representation of the equation [tex]\( h = -16t^2 + 80 \)[/tex].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.