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-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: (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)

Answer 1

To analyze 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 standard form \( ax^2 + bx + c \) can be calculated using the formula:
\[
x = -\frac{b}{2a}
\]
For \( y = x^2 - 2x - 120 \):
- \( a = 1 \)
- \( b = -2 \)

Calculating the x-coordinate of the vertex:
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]

Substituting \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
Thus, the vertex is \( (1, -121) \).

### 2. **x-intercepts**
To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
Factoring the quadratic:
\[
0 = (x - 12)(x + 10)
\]
Setting each factor to zero gives us the x-intercepts:
- \( x - 12 = 0 \) → \( x = 12 \)
- \( x + 10 = 0 \) → \( x = -10 \)

Therefore, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).

### 3. **y-intercept**
To find the y-intercept, set \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
Thus, 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
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)