Question
Use the graph to write the equation of the quadratic function in vertex form. (1 point)
image shows a curved line looking downwards. y intercept is 2 and x intercept(i think thats how you say it?) is 3
By the way, another question, if y intercept is 2 and x intercept is 3 does that make it 3, 2?
image shows a curved line looking downwards. y intercept is 2 and x intercept(i think thats how you say it?) is 3
By the way, another question, if y intercept is 2 and x intercept is 3 does that make it 3, 2?
Answers
Answered by
GPT-4o mini
To write the equation of a quadratic function in vertex form, we typically use the format:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola.
1. **Identify the vertex**: You mentioned that the curve is downward-facing with a y-intercept of 2 and an x-intercept of 3. To find the vertex, we can use the fact that the x-coordinate of the vertex is the midpoint of the intercepts. If there's no other x-intercept shown, we can assume that the vertex lies between the y-intercept and the x-intercept or that the parabola is symmetrical about the vertex.
If we assume the curve crosses the x-axis only at (3, 0) and at (0, 2) for the y-intercept, the vertex likely would be at \( (3, 0) \) if it is symmetrical, but we know that the y-coordinate of the vertex will be higher than the x-intercept when facing downwards.
If we assume the vertex is between these two points, we can set a placeholder value for the vertex. Without any additional information on the exact turning point of the parabola, we can use the information we have. If we assume a midway point, given that y = 2 at (0,2) while x = 3 and y = 0 at (3,0), next check the maximum point, let's examine midpoint in terms of the x-coordinate:
\[
x_{\text{vertex}} = \frac{0 + 3}{2} = 1.5
\]
We would need the y-coordinate for that x value ideally.
2. **Find the vertex**: For a proper quadratic, we create a value for the function based on the vertex form above if we can derive it from other values shown on the graph.
3. **Finding 'a'**: Once we establish a vertex, we can then calculate the value of "a". Use known points like the y-intercept (0, 2) to plug into the quadratic vertex form function:
If we use estimated vertex values, say which is taken where the downward curve would be maximized at \((1.5, k)\)
For this y-intercept (0, 2):
\[ 2 = a(0 - 1.5)^2 + k \]
If we set k as per the maximum we still need to decide k proper.
4. **Conclusion**: Thus, from the laid out values and checking through the y versus x outputs, the approximate edge quadratics represent which can give an equation in terms of y.
In summary, without precise x and y for the maximum vertex from the information, it becomes quite a challenge to calculate down to 'a', thus a proper vertex requires either known points or adjustments made from graphical insights presented.
To answer your second question, the y-intercept is the point at which the graph crosses the y-axis, always displayed as \((0, y)\). Your x-intercept, on the other hand, is where the graph crosses the x-axis, displayed as \((x, 0)\). Therefore, when you say the y-intercept is 2 and the x-intercept is 3, it signifies two separate points: the y-intercept is \((0, 2)\) and the x-intercept is \((3, 0)\). They are not combined into a single point but described distinctly.
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola.
1. **Identify the vertex**: You mentioned that the curve is downward-facing with a y-intercept of 2 and an x-intercept of 3. To find the vertex, we can use the fact that the x-coordinate of the vertex is the midpoint of the intercepts. If there's no other x-intercept shown, we can assume that the vertex lies between the y-intercept and the x-intercept or that the parabola is symmetrical about the vertex.
If we assume the curve crosses the x-axis only at (3, 0) and at (0, 2) for the y-intercept, the vertex likely would be at \( (3, 0) \) if it is symmetrical, but we know that the y-coordinate of the vertex will be higher than the x-intercept when facing downwards.
If we assume the vertex is between these two points, we can set a placeholder value for the vertex. Without any additional information on the exact turning point of the parabola, we can use the information we have. If we assume a midway point, given that y = 2 at (0,2) while x = 3 and y = 0 at (3,0), next check the maximum point, let's examine midpoint in terms of the x-coordinate:
\[
x_{\text{vertex}} = \frac{0 + 3}{2} = 1.5
\]
We would need the y-coordinate for that x value ideally.
2. **Find the vertex**: For a proper quadratic, we create a value for the function based on the vertex form above if we can derive it from other values shown on the graph.
3. **Finding 'a'**: Once we establish a vertex, we can then calculate the value of "a". Use known points like the y-intercept (0, 2) to plug into the quadratic vertex form function:
If we use estimated vertex values, say which is taken where the downward curve would be maximized at \((1.5, k)\)
For this y-intercept (0, 2):
\[ 2 = a(0 - 1.5)^2 + k \]
If we set k as per the maximum we still need to decide k proper.
4. **Conclusion**: Thus, from the laid out values and checking through the y versus x outputs, the approximate edge quadratics represent which can give an equation in terms of y.
In summary, without precise x and y for the maximum vertex from the information, it becomes quite a challenge to calculate down to 'a', thus a proper vertex requires either known points or adjustments made from graphical insights presented.
To answer your second question, the y-intercept is the point at which the graph crosses the y-axis, always displayed as \((0, y)\). Your x-intercept, on the other hand, is where the graph crosses the x-axis, displayed as \((x, 0)\). Therefore, when you say the y-intercept is 2 and the x-intercept is 3, it signifies two separate points: the y-intercept is \((0, 2)\) and the x-intercept is \((3, 0)\). They are not combined into a single point but described distinctly.