To find the key points of the quadratic function \( y = x^2 - 2x - 120 \), we'll determine the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a quadratic in standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -2 \).
\[ x = -\frac{-2}{2 \cdot 1} = 1 \]
To find the y-coordinate of the vertex, substitute \( x = 1 \) back into the equation:
\[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]
Thus, the vertex is (1, -121).
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X-intercepts: To find the x-intercepts, set \( y = 0 \):
\[ 0 = x^2 - 2x - 120 \]
Factoring the quadratic gives:
\[ (x - 12)(x + 10) = 0 \]
Thus, the x-intercepts are:
\[ x - 12 = 0 \Rightarrow x = 12 \quad \text{and} \quad x + 10 = 0 \Rightarrow x = -10 \]
Therefore, the x-intercepts are (-10, 0) and (12, 0).
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Y-intercept: The y-intercept is found by evaluating \( y \) at \( x = 0 \):
\[ y = (0)^2 - 2(0) - 120 = -120 \]
Therefore, the y-intercept is (0, -120).
Summary of Key Points:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
The correct option from your provided responses would be:
x-intercepts: (-10, 0) (12, 0); y-intercept: (0, -120); vertex: (1, -121).