Question

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-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

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

Answer 1

To find the key points on the graph of the quadratic function \(y = x^2 - 2x - 120\), we can calculate 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:
\[
x = -\frac{b}{2a}
\]
For the given function, \(a = 1\) and \(b = -2\):
\[
x = -\frac{-2}{2 \cdot 1} = 1
\]
Now substituting \(x = 1\) back into the function to find \(y\):
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
So the vertex is \((1, -121)\).

2. **Y-Intercept**: The y-intercept occurs when \(x = 0\):
\[
y = 0^2 - 2(0) - 120 = -120
\]
Hence, the y-intercept is \((0, -120)\).

3. **X-Intercepts**: The x-intercepts occur where \(y = 0\):
\[
0 = x^2 - 2x - 120
\]
Factoring the quadratic gives:
\[
(x - 12)(x + 10) = 0
\]
Therefore, the x-intercepts are \(x = 12\) and \(x = -10\), leading to the points \((12, 0)\) and \((-10, 0)\).

**Summary of Key Points**:
- **Vertex**: \((1, -121)\)
- **X-Intercepts**: \((-10, 0)\) and \((12, 0)\)
- **Y-Intercept**: \((0, -120)\)

**Final Answer**:
- x-intercepts: \((-10, 0)\), \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)