Question
Using the vertex (0.5,9) the axis of symmetry (0.5), the y intercept (0,5) and the y intercept (1.25,0) provide a real life example
Answers
Answer
Sorry the (1.25,0) is the x intercept
Answer
Can you please show me the equations
Answered by
GPT-4o mini
Let's create a real-life example that utilizes the mathematical concepts of vertex, axis of symmetry, and intercepts from a quadratic function, particularly in the context of projectile motion.
### Example: Water Fountain Design
Imagine you are an engineer tasked with designing a decorative water fountain in a city park. The fountain will shoot water in a parabolic arc. You want to ensure that it has a pleasing visual effect while also being functional and efficient in terms of water usage.
### Key Features of the Fountain's Arc:
1. **Vertex (0.5, 9)**: The vertex of the parabola represents the highest point of the water arc, which is 9 units high at the horizontal distance of 0.5 meters from the fountain's base. This represents the apex of the water trajectory, where it is the most visually striking.
2. **Axis of Symmetry (0.5)**: The axis of symmetry being at x = 0.5 means that the water arcs equally on both sides of this point. This symmetry ensures that the fountain has an aesthetically balanced look, as the water will shoot out and fall symmetrically around the center point of the fountain.
3. **Y-Intercept (0, 5)**: The y-intercept indicates that when the horizontal distance is 0 (at the base of the fountain), the height of the water begins at 5 units. This indicates the initial, static height of the water before it shoots upwards.
4. **Another Y-Intercept (1.25, 0)**: This point indicates where the water touches back down to the ground level as it completes its arc. At a horizontal distance of 1.25 meters from the base of the fountain, the water reaches the ground, making it clear where the arc ends and aiding in placement considerations for surrounding vegetation or seating.
### Application:
The water fountain, designed with these parameters, enhances the visual appeal of the park, attracting visitors and creating a refreshing environment. Additionally, knowing the height and distances involved helps in planning maintenance access and ensuring safety regulations around the fountain.
Thus, through the study of the quadratic function that defines the water's arc, the engineer can create an engaging feature that is not only functional but also enhances the park's ambiance.
### Example: Water Fountain Design
Imagine you are an engineer tasked with designing a decorative water fountain in a city park. The fountain will shoot water in a parabolic arc. You want to ensure that it has a pleasing visual effect while also being functional and efficient in terms of water usage.
### Key Features of the Fountain's Arc:
1. **Vertex (0.5, 9)**: The vertex of the parabola represents the highest point of the water arc, which is 9 units high at the horizontal distance of 0.5 meters from the fountain's base. This represents the apex of the water trajectory, where it is the most visually striking.
2. **Axis of Symmetry (0.5)**: The axis of symmetry being at x = 0.5 means that the water arcs equally on both sides of this point. This symmetry ensures that the fountain has an aesthetically balanced look, as the water will shoot out and fall symmetrically around the center point of the fountain.
3. **Y-Intercept (0, 5)**: The y-intercept indicates that when the horizontal distance is 0 (at the base of the fountain), the height of the water begins at 5 units. This indicates the initial, static height of the water before it shoots upwards.
4. **Another Y-Intercept (1.25, 0)**: This point indicates where the water touches back down to the ground level as it completes its arc. At a horizontal distance of 1.25 meters from the base of the fountain, the water reaches the ground, making it clear where the arc ends and aiding in placement considerations for surrounding vegetation or seating.
### Application:
The water fountain, designed with these parameters, enhances the visual appeal of the park, attracting visitors and creating a refreshing environment. Additionally, knowing the height and distances involved helps in planning maintenance access and ensuring safety regulations around the fountain.
Thus, through the study of the quadratic function that defines the water's arc, the engineer can create an engaging feature that is not only functional but also enhances the park's ambiance.
Answered by
GPT-4o mini
No problem! Let's clarify how the points relate to the quadratic function based on the correct information, including the \( (1.25, 0) \) as an x-intercept (also known as the root of the function).
### Real-life Example: Water Fountain Design (Revised)
Imagine you are an engineer tasked with designing a decorative water fountain in a city park. The fountain will shoot water in a parabolic arc. You want to ensure that it has a pleasing visual effect while also being functional and efficient in terms of water usage.
### Key Features of the Fountain's Arc and Quadratic Function:
1. **Vertex (0.5, 9)**: The vertex of the parabola represents the highest point of the water arc, which is 9 units high at the horizontal distance of 0.5 meters from the fountain's base. This indicates the apex of the water trajectory, where it is the most visually striking and impressive.
2. **Axis of Symmetry (0.5)**: The axis of symmetry being at \( x = 0.5 \) means that the water arcs equally on both sides of this point. This symmetry ensures that the fountain has an aesthetically balanced look, as the water will shoot out and fall symmetrically around the center point of the fountain.
3. **Y-Intercept (0, 5)**: The y-intercept indicates that when the horizontal distance is 0 (at the base of the fountain), the height of the water begins at 5 units. This represents the static height of the water before it shoots upwards. This means that when the water is still, it sits at 5 units above ground level.
4. **X-Intercept (1.25, 0)**: This point indicates where the water touches back down to the ground level as it completes its arc. At a horizontal distance of 1.25 meters from the base of the fountain, the water reaches the ground, marking the end of its trajectory. This can be crucial for ensuring the surrounding area is designed appropriately for when the water lands, such as ensuring that it's directed away from areas where people might walk.
### Application:
The water fountain, designed with these parameters, enhances the visual appeal of the park, attracting visitors and creating a refreshing and enjoyable environment. Knowing the height, distances, and intercepts involved aids in planning maintenance access, positioning seating areas, or landscaping around the fountain to avoid excess splashing.
Using this curvature, the fountain can provide a captivating display, with water shooting high before gracefully descending back to the ground, ensuring that it is not only functional but also a beautiful focal point for park-goers. This understanding of the parabolic trajectory allows the engineer to optimize the performance and aesthetics of the fountain effectively.
### Real-life Example: Water Fountain Design (Revised)
Imagine you are an engineer tasked with designing a decorative water fountain in a city park. The fountain will shoot water in a parabolic arc. You want to ensure that it has a pleasing visual effect while also being functional and efficient in terms of water usage.
### Key Features of the Fountain's Arc and Quadratic Function:
1. **Vertex (0.5, 9)**: The vertex of the parabola represents the highest point of the water arc, which is 9 units high at the horizontal distance of 0.5 meters from the fountain's base. This indicates the apex of the water trajectory, where it is the most visually striking and impressive.
2. **Axis of Symmetry (0.5)**: The axis of symmetry being at \( x = 0.5 \) means that the water arcs equally on both sides of this point. This symmetry ensures that the fountain has an aesthetically balanced look, as the water will shoot out and fall symmetrically around the center point of the fountain.
3. **Y-Intercept (0, 5)**: The y-intercept indicates that when the horizontal distance is 0 (at the base of the fountain), the height of the water begins at 5 units. This represents the static height of the water before it shoots upwards. This means that when the water is still, it sits at 5 units above ground level.
4. **X-Intercept (1.25, 0)**: This point indicates where the water touches back down to the ground level as it completes its arc. At a horizontal distance of 1.25 meters from the base of the fountain, the water reaches the ground, marking the end of its trajectory. This can be crucial for ensuring the surrounding area is designed appropriately for when the water lands, such as ensuring that it's directed away from areas where people might walk.
### Application:
The water fountain, designed with these parameters, enhances the visual appeal of the park, attracting visitors and creating a refreshing and enjoyable environment. Knowing the height, distances, and intercepts involved aids in planning maintenance access, positioning seating areas, or landscaping around the fountain to avoid excess splashing.
Using this curvature, the fountain can provide a captivating display, with water shooting high before gracefully descending back to the ground, ensuring that it is not only functional but also a beautiful focal point for park-goers. This understanding of the parabolic trajectory allows the engineer to optimize the performance and aesthetics of the fountain effectively.
Answered by
GPT-4o mini
Certainly! To create a quadratic equation based on the conditions you've provided—specifically the vertex, x-intercept, and y-intercept—we can use the general form of a quadratic equation:
\[
y = ax^2 + bx + c
\]
To derive this equation, we can use the Vertex Form of a quadratic equation, which is:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola. From your data, the vertex is \((0.5, 9)\), so we can plug in these values:
\[
y = a(x - 0.5)^2 + 9
\]
### Step 1: Finding \( a \)
To find the value of \( a \), we can use the x-intercept \((1.25, 0)\). An x-intercept occurs when \( y = 0 \), so we can set \( y = 0 \) and \( x = 1.25\):
\[
0 = a(1.25 - 0.5)^2 + 9
\]
This simplifies to:
\[
0 = a(0.75)^2 + 9
\]
\[
0 = 0.5625a + 9
\]
Now, solving for \( a \):
\[
0.5625a = -9
\]
\[
a = \frac{-9}{0.5625} = -16
\]
### Step 2: Write the Equation in Vertex Form
Now that we have \( a \), we can write the equation:
\[
y = -16(x - 0.5)^2 + 9
\]
### Step 3: Convert to Standard Form
To convert this to standard form, we can expand it:
1. Expand the squared term:
\[
y = -16(x^2 - 1.0x + 0.25) + 9
\]
\[
y = -16x^2 + 16x - 4 + 9
\]
2. Combine like terms:
\[
y = -16x^2 + 16x + 5
\]
### Summary of the Quadratic Equation
So, the equation of the parabola in standard form is:
\[
y = -16x^2 + 16x + 5
\]
### Verification of Key Points
1. **Vertex**: We can check that the vertex \((0.5, 9)\) lies on this curve by plugging \( x = 0.5 \) into the equation:
\[
y = -16(0.5)^2 + 16(0.5) + 5
\]
\[
y = -16(0.25) + 8 + 5 = -4 + 8 + 5 = 9
\]
2. **Y-Intercept**: Check \( (0,5) \):
\[
y = -16(0)^2 + 16(0) + 5 = 5
\]
3. **X-Intercept**: Check \( (1.25,0) \):
\[
y = -16(1.25)^2 + 16(1.25) + 5
\]
\[
y = -16(1.5625) + 20 + 5 = -25 + 20 + 5 = 0
\]
All key points check out correctly! Thus, the quadratic function representing the fountain arc is:
\[
y = -16x^2 + 16x + 5
\]
\[
y = ax^2 + bx + c
\]
To derive this equation, we can use the Vertex Form of a quadratic equation, which is:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola. From your data, the vertex is \((0.5, 9)\), so we can plug in these values:
\[
y = a(x - 0.5)^2 + 9
\]
### Step 1: Finding \( a \)
To find the value of \( a \), we can use the x-intercept \((1.25, 0)\). An x-intercept occurs when \( y = 0 \), so we can set \( y = 0 \) and \( x = 1.25\):
\[
0 = a(1.25 - 0.5)^2 + 9
\]
This simplifies to:
\[
0 = a(0.75)^2 + 9
\]
\[
0 = 0.5625a + 9
\]
Now, solving for \( a \):
\[
0.5625a = -9
\]
\[
a = \frac{-9}{0.5625} = -16
\]
### Step 2: Write the Equation in Vertex Form
Now that we have \( a \), we can write the equation:
\[
y = -16(x - 0.5)^2 + 9
\]
### Step 3: Convert to Standard Form
To convert this to standard form, we can expand it:
1. Expand the squared term:
\[
y = -16(x^2 - 1.0x + 0.25) + 9
\]
\[
y = -16x^2 + 16x - 4 + 9
\]
2. Combine like terms:
\[
y = -16x^2 + 16x + 5
\]
### Summary of the Quadratic Equation
So, the equation of the parabola in standard form is:
\[
y = -16x^2 + 16x + 5
\]
### Verification of Key Points
1. **Vertex**: We can check that the vertex \((0.5, 9)\) lies on this curve by plugging \( x = 0.5 \) into the equation:
\[
y = -16(0.5)^2 + 16(0.5) + 5
\]
\[
y = -16(0.25) + 8 + 5 = -4 + 8 + 5 = 9
\]
2. **Y-Intercept**: Check \( (0,5) \):
\[
y = -16(0)^2 + 16(0) + 5 = 5
\]
3. **X-Intercept**: Check \( (1.25,0) \):
\[
y = -16(1.25)^2 + 16(1.25) + 5
\]
\[
y = -16(1.5625) + 20 + 5 = -25 + 20 + 5 = 0
\]
All key points check out correctly! Thus, the quadratic function representing the fountain arc is:
\[
y = -16x^2 + 16x + 5
\]
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.