To find the key points of the quadratic equation \( y = x^2 - 2x - 120 \), we can follow these steps:
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Find the vertex: The vertex form of a quadratic equation is given by the formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = -2 \).
\[ x = -\frac{-2}{2(1)} = 1 \]
To find the y-coordinate of the vertex, substitute \( x = 1 \) back into the original equation:
\[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]
So, the vertex is at \( (1, -121) \).
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Find the x-intercepts: Set \( y = 0 \):
\[ 0 = x^2 - 2x - 120 \]
To solve for \( x \), we can factor the quadratic:
\[ 0 = (x - 12)(x + 10) \]
Setting each factor to zero gives the solutions:
\[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x + 10 = 0 \quad \Rightarrow \quad x = -10 \]
So the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
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Find the y-intercept: Set \( x = 0 \):
\[ y = (0)^2 - 2(0) - 120 = -120 \]
So the y-intercept is \( (0, -120) \).
Summary:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)
None of the provided responses contain the correct combination of vertex, x-intercepts, and y-intercept. The correct key points are:
- Vertex: \( (1, -121) \)
- X-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- Y-intercept: \( (0, -120) \)