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: (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-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)

1 answer

To find the key points on the graph of the quadratic equation \( y = x^2 - 2x - 120 \), we need to identify the vertex, x-intercepts, and y-intercept.

  1. Vertex: The vertex of a parabola represented by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). In this case, \( a = 1 \) and \( b = -2 \). \[ x = -\frac{-2}{2 \cdot 1} = 1 \] Substitute \( x = 1 \) back into the equation 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 \( x = 0 \). \[ y = (0)^2 - 2(0) - 120 = -120 \] Therefore, the y-intercept is \( (0, -120) \).

  3. X-intercepts: The x-intercepts are found by setting \( y = 0 \) in the equation: \[ 0 = x^2 - 2x - 120 \] Factoring the quadratic gives: \[ 0 = (x - 12)(x + 10) \] Thus, the solutions are: \[ x - 12 = 0 \Rightarrow x = 12 \quad \text{and} \quad x + 10 = 0 \Rightarrow x = -10 \] This means the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).

Putting it all together, here are the key points:

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

Thus, the correct response is:

  • x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
  • y-intercept: \( (0, -120) \)
  • vertex: \( (1, -121) \)