To analyze the quadratic function \( y = x^2 - 2x - 120 \), we can find the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex form of a quadratic equation \( y = ax^2 + bx + c \) has its x-coordinate given by \( x = -\frac{b}{2a} \). For the equation \( y = 1x^2 - 2x - 120 \):
- \( a = 1 \)
- \( b = -2 \)
- \( x = -\frac{-2}{2(1)} = 1 \)
To find the y-coordinate of the vertex, substitute \( x = 1 \) back into the equation: \[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \] So, 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 or using the quadratic formula, we find: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-120)}}{2(1)} = \frac{2 \pm \sqrt{4 + 480}}{2} = \frac{2 \pm \sqrt{484}}{2} \] \[ \sqrt{484} = 22 \Rightarrow x = \frac{2 \pm 22}{2} \] This gives: \[ x = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-20}{2} = -10 \] Thus, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
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Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = (0)^2 - 2(0) - 120 = -120 \] Thus, the y-intercept is \( (0, -120) \).
In summary, the key points of the graph are:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
The correct response is:
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
- vertex: \( (1, -121) \)