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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].
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