Question
The vertex of a parabola is located at (1, -7). The parabola intersects the x-axis between 6 and 7.
Between which two negative integers will the parabola intersect the x-axis?
• A. between -8 and -7
B. between -7 and -6
• c. between -6 and -5
D. between -5 and -4
E. between -4 and -3
Between which two negative integers will the parabola intersect the x-axis?
• A. between -8 and -7
B. between -7 and -6
• c. between -6 and -5
D. between -5 and -4
E. between -4 and -3
Answers
GPT-4o mini
To determine the x-intercepts of the parabola, we can use the information about its vertex and its intersection with the x-axis.
The vertex of the parabola is given as \( (1, -7) \). Since the parabola opens upwards or downwards, the general form of a quadratic equation can be expressed as:
\[
y = a(x - 1)^2 - 7
\]
To find the value of \( a \), we need additional information. We know the parabola intersects the x-axis between 6 and 7. Thus, both points \( (6, 0) \) and \( (7, 0) \) will lead us to determine \( a \).
Let's first express the points where the parabola intersects the x-axis using the parabola equation:
1. **At \( x = 6 \):**
\[
0 = a(6 - 1)^2 - 7
\]
\[
0 = 25a - 7
\]
\[
25a = 7 \quad \Rightarrow \quad a = \frac{7}{25}
\]
2. **At \( x = 7 \):**
\[
0 = a(7 - 1)^2 - 7
\]
\[
0 = 36a - 7
\]
\[
36a = 7 \quad \Rightarrow \quad a = \frac{7}{36}
\]
However, since both conditions cannot be satisfied with a single value of \( a \), it must imply that the parabola intersects at another point on the x-axis besides x = 6 and x = 7, likely due to symmetry around the vertex.
### Finding X-Intercepts
To find the x-intercepts of the parabola further, we know that from the vertex at (1, -7), the parabola will also be symmetric and we can set it to \( y = 0 \):
\[
0 = a(x - 1)^2 - 7
\]
\[
a(x - 1)^2 = 7
\]
\[
(x - 1)^2 = \frac{7}{a}
\]
\[
x - 1 = \pm\sqrt{\frac{7}{a}}
\]
\[
x = 1 \pm \sqrt{\frac{7}{a}}
\]
Given that we’ll deal with increasing values of \( x \) because of previous information. Calculating further simplifies that the critical point would reflect in negative integers around the vertex towards negative integer roots.
The value of \( a \) ultimately determines the bowl's opening direction. Since 7 is a positive value, we can assume it intersects the x-axis on the negative side.
By testing negative integers, we need to consider \( (1 ± \sqrt{\frac{7}{a}}) \) will likely fall between two journeys around the vertex to validate the equal roots.
Checking through choices:
- **A. between -8 and -7** (possible)
- **B. between -7 and -6** (possible)
- **C. between -6 and -5** (possible)
- **D. between -5 and -4** (possible)
- **E. between -4 and -3** (possible)
Since this parabola opens upward down the line, projecting higher roots add up. The intersective scope of analysis mostly points roots within -8 and -6 consequently as they ascend towards (1, -7).
Thus, the negative integers that the parabola intersects the x-axis lie between:
**Answer is B. between -7 and -6.**
The vertex of the parabola is given as \( (1, -7) \). Since the parabola opens upwards or downwards, the general form of a quadratic equation can be expressed as:
\[
y = a(x - 1)^2 - 7
\]
To find the value of \( a \), we need additional information. We know the parabola intersects the x-axis between 6 and 7. Thus, both points \( (6, 0) \) and \( (7, 0) \) will lead us to determine \( a \).
Let's first express the points where the parabola intersects the x-axis using the parabola equation:
1. **At \( x = 6 \):**
\[
0 = a(6 - 1)^2 - 7
\]
\[
0 = 25a - 7
\]
\[
25a = 7 \quad \Rightarrow \quad a = \frac{7}{25}
\]
2. **At \( x = 7 \):**
\[
0 = a(7 - 1)^2 - 7
\]
\[
0 = 36a - 7
\]
\[
36a = 7 \quad \Rightarrow \quad a = \frac{7}{36}
\]
However, since both conditions cannot be satisfied with a single value of \( a \), it must imply that the parabola intersects at another point on the x-axis besides x = 6 and x = 7, likely due to symmetry around the vertex.
### Finding X-Intercepts
To find the x-intercepts of the parabola further, we know that from the vertex at (1, -7), the parabola will also be symmetric and we can set it to \( y = 0 \):
\[
0 = a(x - 1)^2 - 7
\]
\[
a(x - 1)^2 = 7
\]
\[
(x - 1)^2 = \frac{7}{a}
\]
\[
x - 1 = \pm\sqrt{\frac{7}{a}}
\]
\[
x = 1 \pm \sqrt{\frac{7}{a}}
\]
Given that we’ll deal with increasing values of \( x \) because of previous information. Calculating further simplifies that the critical point would reflect in negative integers around the vertex towards negative integer roots.
The value of \( a \) ultimately determines the bowl's opening direction. Since 7 is a positive value, we can assume it intersects the x-axis on the negative side.
By testing negative integers, we need to consider \( (1 ± \sqrt{\frac{7}{a}}) \) will likely fall between two journeys around the vertex to validate the equal roots.
Checking through choices:
- **A. between -8 and -7** (possible)
- **B. between -7 and -6** (possible)
- **C. between -6 and -5** (possible)
- **D. between -5 and -4** (possible)
- **E. between -4 and -3** (possible)
Since this parabola opens upward down the line, projecting higher roots add up. The intersective scope of analysis mostly points roots within -8 and -6 consequently as they ascend towards (1, -7).
Thus, the negative integers that the parabola intersects the x-axis lie between:
**Answer is B. between -7 and -6.**