To determine the key points of the quadratic function \(y = x^2 - 2x - 120\), we can follow these steps:
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Vertex: The vertex of a quadratic function \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] Here, \(a = 1\) and \(b = -2\).
\[ x = -\frac{-2}{2 \times 1} = \frac{2}{2} = 1 \]
To find the y-coordinate of the vertex, we substitute \(x = 1\) back into the equation: \[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]
Thus, the vertex is: \[ \text{vertex: } (1, -121) \]
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Y-intercept: The y-intercept can be found by evaluating \(y\) when \(x = 0\): \[ y = (0)^2 - 2(0) - 120 = -120 \]
Therefore, the y-intercept is: \[ \text{y-intercept: } (0, -120) \]
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X-intercepts: The x-intercepts occur when \(y = 0\): \[ 0 = x^2 - 2x - 120 \] We can factor this quadratic equation or use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Here, \( a = 1 \), \( b = -2 \), and \( c = -120 \).
The discriminant \(D\) is: \[ D = b^2 - 4ac = (-2)^2 - 4(1)(-120) = 4 + 480 = 484 \]
Now applying the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{484}}{2(1)} = \frac{2 \pm 22}{2} \] This gives us two solutions: \[ x_1 = \frac{24}{2} = 12 \quad \text{(rightmost intercept)} \] \[ x_2 = \frac{-20}{2} = -10 \quad \text{(leftmost intercept)} \]
Therefore, the x-intercepts are: \[ \text{x-intercepts: } (-10, 0) \text{ and } (12, 0) \]
Summary of key points:
- x-intercepts: \((-10, 0)\) and \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)
Please note to use the x-intercepts in the format requested by providing the leftmost point first.