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-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: (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-intercept: (1,−121)

y-intercept: (0,−120)

vertex: (−10,0) (12,0)

Answer 1

To find the key points of the quadratic equation \( y = x^2 - 2x - 120 \), we can follow these steps:

1. **Find the vertex:**
The vertex form of a quadratic equation is given by the formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = -2 \).

\[
x = -\frac{-2}{2(1)} = 1
\]

To find the y-coordinate of the vertex, substitute \( x = 1 \) back into the original equation:

\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]

So, the vertex is at \( (1, -121) \).

2. **Find the x-intercepts:**
Set \( y = 0 \):

\[
0 = x^2 - 2x - 120
\]

To solve for \( x \), we can factor the quadratic:

\[
0 = (x - 12)(x + 10)
\]

Setting each factor to zero gives the solutions:

\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x + 10 = 0 \quad \Rightarrow \quad x = -10
\]

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

3. **Find the y-intercept:**
Set \( x = 0 \):

\[
y = (0)^2 - 2(0) - 120 = -120
\]

So the y-intercept is \( (0, -120) \).

### Summary:

- **Vertex:** \( (1, -121) \)
- **X-intercepts:** \( (-10, 0) \) and \( (12, 0) \)
- **Y-intercept:** \( (0, -120) \)

None of the provided responses contain the correct combination of vertex, x-intercepts, and y-intercept. The correct key points are:

- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)