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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: (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-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

1 answer

To find the key points of the quadratic function \( y = x^2 - 2x - 120 \), we can go through the process of identifying the vertex, x-intercepts, and y-intercept.

  1. Vertex: The vertex can be found using the formula \( x = -\frac{b}{2a} \) for the quadratic equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = -2 \).

    \[ x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1 \]

    Now substituting \( x = 1 \) back into the equation to find the y-coordinate:

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

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

  2. X-Intercepts: To find the x-intercepts, set \( y = 0 \):

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

    Factoring the quadratic:

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

    Therefore, the solutions are:

    \[ x = 12 \quad \text{and} \quad x = -10 \]

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

  3. Y-Intercept: To find the y-intercept, set \( x = 0 \):

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

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

Summary of Key Points:

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

Final response:

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