To find the key points on the graph of the quadratic function \(y = x^2 - 2x - 120\), we'll follow these steps:
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Find the x-intercepts: To find the x-intercepts, set \(y = 0\) and solve the equation \(x^2 - 2x - 120 = 0\).
The equation can be factored as follows:
\[ x^2 - 2x - 120 = (x - 12)(x + 10) = 0 \]
Setting each factor to zero gives:
\[ x - 12 = 0 \implies x = 12 \] \[ x + 10 = 0 \implies x = -10 \]
Therefore, the x-intercepts are: \[ \text{x-intercepts: } (-10, 0) \text{ and } (12, 0) \]
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Find the y-intercept: To find the y-intercept, set \(x = 0\) in the original equation:
\[ y = 0^2 - 2(0) - 120 = -120 \]
So, the y-intercept is: \[ \text{y-intercept: } (0, -120) \]
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Find the vertex: The vertex of a parabola in standard form \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} \] Here, \(a = 1\) and \(b = -2\):
\[ x = -\frac{-2}{2(1)} = \frac{2}{2} = 1 \]
Now substitute \(x = 1\) back into the original equation to find the y-coordinate of the vertex:
\[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]
So, the vertex is: \[ \text{vertex: } (1, -121) \]
In summary, the key points are:
- x-intercepts: \((-10, 0)\) and \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)