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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

1 answer

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: \((1, -121)\)