What are the key points on the graph of y=x2−2x−120? Name the vertex, x-intercept(s), and y-intercept.(1 point)

Question

Responses

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)

Answer 1

To find the key points of the quadratic function \(y = x^2 - 2x - 120\), we can determine 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 \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = -2\).

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

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

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

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

2. Y-intercept: The y-intercept occurs when \(x = 0\):

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

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

3. X-intercepts: The x-intercepts occur when \(y = 0\):

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

   To find the x-intercepts, we can factor the expression:

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

   This gives us the solutions \(x = 12\) and \(x = -10\), thus the x-intercepts are:

   \[ (12, 0) \quad \text{and} \quad (-10, 0) \]

Putting it all together, the key points for the graph of the function \(y = x^2 - 2x - 120\) are:

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

Therefore, both sets of responses you provided about the x-intercepts, y-intercept, and vertex are correct.