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
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
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
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
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
? Name the vertex, x-intercept(s), and y-intercept.(1 point)
Responses
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
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
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
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
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the key points of the quadratic function \( y = x^2 - 2x - 120 \), we can go through the process of identifying the vertex, x-intercepts, and y-intercept.
1. **Vertex**: The vertex can be found using the formula \( x = -\frac{b}{2a} \) for the quadratic equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = -2 \).
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
Now substituting \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So the vertex is \( (1, -121) \).
2. **X-Intercepts**: To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
Factoring the quadratic:
\[
(x - 12)(x + 10) = 0
\]
Therefore, the solutions are:
\[
x = 12 \quad \text{and} \quad x = -10
\]
So the x-intercepts are \( (12, 0) \) and \( (-10, 0) \).
3. **Y-Intercept**: To find the y-intercept, set \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
So the y-intercept is \( (0, -120) \).
### Summary of Key Points:
- **Vertex**: \( (1, -121) \)
- **X-Intercepts**: \( (-10, 0) \) and \( (12, 0) \)
- **Y-Intercept**: \( (0, -120) \)
**Final response**:
- x-intercepts: \( (-10, 0), (12, 0) \)
- y-intercept: \( (0, -120) \)
- vertex: \( (1, -121) \)
1. **Vertex**: The vertex can be found using the formula \( x = -\frac{b}{2a} \) for the quadratic equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = -2 \).
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
Now substituting \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So the vertex is \( (1, -121) \).
2. **X-Intercepts**: To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
Factoring the quadratic:
\[
(x - 12)(x + 10) = 0
\]
Therefore, the solutions are:
\[
x = 12 \quad \text{and} \quad x = -10
\]
So the x-intercepts are \( (12, 0) \) and \( (-10, 0) \).
3. **Y-Intercept**: To find the y-intercept, set \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
So the y-intercept is \( (0, -120) \).
### Summary of Key Points:
- **Vertex**: \( (1, -121) \)
- **X-Intercepts**: \( (-10, 0) \) and \( (12, 0) \)
- **Y-Intercept**: \( (0, -120) \)
**Final response**:
- x-intercepts: \( (-10, 0), (12, 0) \)
- y-intercept: \( (0, -120) \)
- vertex: \( (1, -121) \)
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