Join the IDNLearn.com community and get your questions answered by experts. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
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].
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.
