Asked by x
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-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
x-intercept: (1,−121)
y-intercept: (0,−120)
Responses
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
x-intercept: (1,−121)
y-intercept: (0,−120)
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 determine the vertex, x-intercepts, and y-intercept.
1. **Vertex**:
The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = -2\).
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
To find the y-coordinate of the vertex, substitute \(x = 1\) back into the equation:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So, the vertex is \((1, -121)\).
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**:
The x-intercepts occur when \(y = 0\):
\[
0 = x^2 - 2x - 120
\]
To find the x-intercepts, we can factor the expression:
\[
(x - 12)(x + 10) = 0
\]
This gives us the solutions \(x = 12\) and \(x = -10\), thus the x-intercepts are:
\[
(12, 0) \quad \text{and} \quad (-10, 0)
\]
Putting it all together, the key points for the graph of the function \(y = x^2 - 2x - 120\) are:
- **Vertex**: \((1, -121)\)
- **X-intercepts**: \((-10, 0)\) and \((12, 0)\)
- **Y-intercept**: \((0, -120)\)
Therefore, both sets of responses you provided about the x-intercepts, y-intercept, and vertex are correct.
1. **Vertex**:
The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = -2\).
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
To find the y-coordinate of the vertex, substitute \(x = 1\) back into the equation:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So, the vertex is \((1, -121)\).
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**:
The x-intercepts occur when \(y = 0\):
\[
0 = x^2 - 2x - 120
\]
To find the x-intercepts, we can factor the expression:
\[
(x - 12)(x + 10) = 0
\]
This gives us the solutions \(x = 12\) and \(x = -10\), thus the x-intercepts are:
\[
(12, 0) \quad \text{and} \quad (-10, 0)
\]
Putting it all together, the key points for the graph of the function \(y = x^2 - 2x - 120\) are:
- **Vertex**: \((1, -121)\)
- **X-intercepts**: \((-10, 0)\) and \((12, 0)\)
- **Y-intercept**: \((0, -120)\)
Therefore, both sets of responses you provided about the x-intercepts, y-intercept, and vertex are correct.
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