Question
Solve by graphing the system of equations:
y=3x2
𝑦
=
3
𝑥
2
y=3x+6
𝑦
=
3
𝑥
+
6
(1 point)
Responses
(−2,12)
(
−
2
,
12
)
and (1,3)
(
1
,
3
)
open paren negative 2 comma 12 close paren and open paren 1 comma 3 close paren - no response given
(−1,3)
(
−
1
,
3
)
and (2,12)
(
2
,
12
)
open paren negative 1 comma 3 close paren and open paren 2 comma 12 close paren - no response given
(2,12)
(
2
,
12
)
open paren 2 comma 12 close paren - no response given
No solution.
y=3x2
𝑦
=
3
𝑥
2
y=3x+6
𝑦
=
3
𝑥
+
6
(1 point)
Responses
(−2,12)
(
−
2
,
12
)
and (1,3)
(
1
,
3
)
open paren negative 2 comma 12 close paren and open paren 1 comma 3 close paren - no response given
(−1,3)
(
−
1
,
3
)
and (2,12)
(
2
,
12
)
open paren negative 1 comma 3 close paren and open paren 2 comma 12 close paren - no response given
(2,12)
(
2
,
12
)
open paren 2 comma 12 close paren - no response given
No solution.
Answers
GPT-4o mini
To solve the system of equations by graphing, we need to analyze and graph both equations:
1. **Quadratic Equation**: \( y = 3x^2 \)
2. **Linear Equation**: \( y = 3x + 6 \)
### Graphing the Quadratic Equation
The equation \( y = 3x^2 \) is a parabola that opens upwards with its vertex at the origin (0, 0). The coefficient 3 means it is narrower than the standard parabola \( y = x^2 \).
We can calculate a few points:
- For \( x = -2: y = 3(-2)^2 = 3(4) = 12 \) → Point (-2, 12)
- For \( x = -1: y = 3(-1)^2 = 3(1) = 3 \) → Point (-1, 3)
- For \( x = 0: y = 3(0)^2 = 0 \) → Point (0, 0)
- For \( x = 1: y = 3(1)^2 = 3(1) = 3 \) → Point (1, 3)
- For \( x = 2: y = 3(2)^2 = 3(4) = 12 \) → Point (2, 12)
### Graphing the Linear Equation
The equation \( y = 3x + 6 \) is a straight line with a slope of 3 and a y-intercept of 6.
We can also calculate a few points:
- For \( x = -2: y = 3(-2) + 6 = -6 + 6 = 0 \) → Point (-2, 0)
- For \( x = -1: y = 3(-1) + 6 = -3 + 6 = 3 \) → Point (-1, 3)
- For \( x = 0: y = 3(0) + 6 = 0 + 6 = 6 \) → Point (0, 6)
- For \( x = 1: y = 3(1) + 6 = 3 + 6 = 9 \) → Point (1, 9)
- For \( x = 2: y = 3(2) + 6 = 6 + 6 = 12 \) → Point (2, 12)
### Intersections
Now, we look for the points at which the two graphs intersect.
From the calculated points:
- For \( x = -1, \) both equations give \( y = 3, \) so the point (-1, 3) is an intersection.
- For \( x = 2, \) both equations give \( y = 12, \) so the point (2, 12) is another intersection point.
- At \( x = -2 \) for the quadratic, \( y = 12 \), but the linear equation gives \( y = 0 \), so these do not intersect at (-2, 12).
### Conclusion
The two intersection points are (-1, 3) and (2, 12).
Therefore, the correct answer is:
**(−1, 3) and (2, 12)**.
1. **Quadratic Equation**: \( y = 3x^2 \)
2. **Linear Equation**: \( y = 3x + 6 \)
### Graphing the Quadratic Equation
The equation \( y = 3x^2 \) is a parabola that opens upwards with its vertex at the origin (0, 0). The coefficient 3 means it is narrower than the standard parabola \( y = x^2 \).
We can calculate a few points:
- For \( x = -2: y = 3(-2)^2 = 3(4) = 12 \) → Point (-2, 12)
- For \( x = -1: y = 3(-1)^2 = 3(1) = 3 \) → Point (-1, 3)
- For \( x = 0: y = 3(0)^2 = 0 \) → Point (0, 0)
- For \( x = 1: y = 3(1)^2 = 3(1) = 3 \) → Point (1, 3)
- For \( x = 2: y = 3(2)^2 = 3(4) = 12 \) → Point (2, 12)
### Graphing the Linear Equation
The equation \( y = 3x + 6 \) is a straight line with a slope of 3 and a y-intercept of 6.
We can also calculate a few points:
- For \( x = -2: y = 3(-2) + 6 = -6 + 6 = 0 \) → Point (-2, 0)
- For \( x = -1: y = 3(-1) + 6 = -3 + 6 = 3 \) → Point (-1, 3)
- For \( x = 0: y = 3(0) + 6 = 0 + 6 = 6 \) → Point (0, 6)
- For \( x = 1: y = 3(1) + 6 = 3 + 6 = 9 \) → Point (1, 9)
- For \( x = 2: y = 3(2) + 6 = 6 + 6 = 12 \) → Point (2, 12)
### Intersections
Now, we look for the points at which the two graphs intersect.
From the calculated points:
- For \( x = -1, \) both equations give \( y = 3, \) so the point (-1, 3) is an intersection.
- For \( x = 2, \) both equations give \( y = 12, \) so the point (2, 12) is another intersection point.
- At \( x = -2 \) for the quadratic, \( y = 12 \), but the linear equation gives \( y = 0 \), so these do not intersect at (-2, 12).
### Conclusion
The two intersection points are (-1, 3) and (2, 12).
Therefore, the correct answer is:
**(−1, 3) and (2, 12)**.