To find the key points on the graph of the function \( y = x^2 - 2x - 120 \), we can analyze the equation.
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 this quadratic:
- \( a = 1 \)
- \( b = -2 \)
Plugging in the values: \[ x = -\frac{-2}{2 \times 1} = 1 \] Now, substitute \( x = 1 \) back into the equation to find the y-coordinate of the vertex: \[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \] Thus, the vertex is at the point: \[ \text{Vertex: } (1, -121) \]
2. Y-Intercept:
The y-intercept occurs when \( x = 0 \): \[ y = (0)^2 - 2(0) - 120 = -120 \] So the y-intercept is at: \[ \text{Y-Intercept: } (0, -120) \]
3. X-Intercepts:
To find the x-intercepts, set \( y = 0 \): \[ 0 = x^2 - 2x - 120 \] This can be factored to: \[ (x - 12)(x + 10) = 0 \] Setting each factor to zero gives: \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x + 10 = 0 \quad \Rightarrow \quad x = -10 \] Thus, the x-intercepts are: \[ \text{X-Intercepts: } (-10, 0) \text{ and } (12, 0) \]
Summary of Intercepts and Vertex:
- X-Intercepts: \((-10, 0)\) and \((12, 0)\)
- Y-Intercept: \((0, -120)\)
- Vertex: \((1, -121)\)
Therefore, the correct response is:
- x-intercepts: \((-10, 0)\) and \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)