To find the key points of the quadratic function \( y = x^2 - 2x - 120 \), we need to identify the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a quadratic in the form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here:
- \( a = 1 \)
- \( b = -2 \)
So: \[ x = -\frac{-2}{2 \cdot 1} = 1 \]
To find the y-value of the vertex, substitute \( x = 1 \) back into the equation: \[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \] Thus, the vertex is \( (1, -121) \).
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X-Intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = x^2 - 2x - 120 \] This can be factored as: \[ (x - 12)(x + 10) = 0 \] Therefore, \( x = 12 \) and \( x = -10 \). The x-intercepts are \( (12, 0) \) and \( (-10, 0) \).
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Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = (0)^2 - 2(0) - 120 = -120 \] Therefore, the y-intercept is \( (0, -120) \).
Summary:
- Vertex: \( (1, -121) \)
- X-Intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-Intercept: \( (0, -120) \)
The correct response based on this analysis is:
- X-Intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-Intercept: \( (0, -120) \)
- Vertex: \( (1, -121) \)