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What are the key points on the graph of y=x2−2x−120 ? Name the vertex, x-intercept(s), and y-intercept.(1 point) Responses x-in...Question
What are the key points on the graph of y=x^2−2x−120
? Name the vertex, x-intercept(s), and y-intercept.(1 point)
Responses
A) x-intercepts: (−10,0)
(12,0)
y-intercept: (1,−121)
vertex: (0,−120)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y-intercept: left parenthesis 1 comma negative 121 right parenthesis vertex: left parenthesis 0 comma negative 120 right parenthesis
B) x-intercepts: (−10,0)
(12,0)
y-intercept: (0,−120)
vertex: (1,−121)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis
C) x-intercept: (0,−120)
y-intercept: (−10,0)
(12,0)
vertex: (1,−121)
x -intercept: left parenthesis 0 comma negative 120 right parenthesis y -intercept: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis
D) x-intercept: (1,−121)
y-intercept: (0,−120)
vertex: (−10,0)
(12,0)
x -intercept: left parenthesis 1 comma negative 121 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis
? Name the vertex, x-intercept(s), and y-intercept.(1 point)
Responses
A) x-intercepts: (−10,0)
(12,0)
y-intercept: (1,−121)
vertex: (0,−120)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y-intercept: left parenthesis 1 comma negative 121 right parenthesis vertex: left parenthesis 0 comma negative 120 right parenthesis
B) x-intercepts: (−10,0)
(12,0)
y-intercept: (0,−120)
vertex: (1,−121)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis
C) x-intercept: (0,−120)
y-intercept: (−10,0)
(12,0)
vertex: (1,−121)
x -intercept: left parenthesis 0 comma negative 120 right parenthesis y -intercept: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis
D) x-intercept: (1,−121)
y-intercept: (0,−120)
vertex: (−10,0)
(12,0)
x -intercept: left parenthesis 1 comma negative 121 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis
Answers
Answered by
GPT-4o mini
To find the key points on the graph of the quadratic function \( y = x^2 - 2x - 120 \), we can analyze the function.
1. **Vertex**:
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -2 \).
\[
x = -\frac{-2}{2 \cdot 1} = 1
\]
Substitute \( x = 1 \) back into the equation to find \( y \):
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So, the vertex is at \( (1, -121) \).
2. **Y-intercept**:
The y-intercept occurs when \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
So, the y-intercept is \( (0, -120) \).
3. **X-intercepts**:
X-intercepts occur where \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
We can factor the quadratic:
\[
0 = (x - 12)(x + 10)
\]
Setting each factor to zero gives:
\[
x - 12 = 0 \implies x = 12 \quad \text{and} \quad x + 10 = 0 \implies x = -10
\]
Therefore, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
Now, summarizing these points:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
The correct response that matches this information is:
**B)**
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
- Vertex: \( (1, -121) \)
1. **Vertex**:
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -2 \).
\[
x = -\frac{-2}{2 \cdot 1} = 1
\]
Substitute \( x = 1 \) back into the equation to find \( y \):
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So, the vertex is at \( (1, -121) \).
2. **Y-intercept**:
The y-intercept occurs when \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
So, the y-intercept is \( (0, -120) \).
3. **X-intercepts**:
X-intercepts occur where \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
We can factor the quadratic:
\[
0 = (x - 12)(x + 10)
\]
Setting each factor to zero gives:
\[
x - 12 = 0 \implies x = 12 \quad \text{and} \quad x + 10 = 0 \implies x = -10
\]
Therefore, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
Now, summarizing these points:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
The correct response that matches this information is:
**B)**
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
- Vertex: \( (1, -121) \)
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