To analyze the quadratic function \( y = x^2 - 2x - 120 \), we can find the key points: the vertex, x-intercepts, and y-intercept.
1. Y-Intercept
The y-intercept is found by setting \( x = 0 \):
\[ y = (0)^2 - 2(0) - 120 = -120 \]
So, the y-intercept is: \[ \text{y-intercept: } (0, -120) \]
2. X-Intercepts
To find the x-intercepts, we set \( y = 0 \):
\[ 0 = x^2 - 2x - 120 \]
This is a quadratic equation, which we can solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = -120 \):
- Calculate the discriminant \( b^2 - 4ac \):
\[ (-2)^2 - 4(1)(-120) = 4 + 480 = 484 \]
- Now apply the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{484}}{2(1)} = \frac{2 \pm 22}{2} \]
This gives:
\[ x_1 = \frac{24}{2} = 12 \] \[ x_2 = \frac{-20}{2} = -10 \]
Thus, the x-intercepts are:
\[ \text{x-intercepts: } (-10, 0) \text{ and } (12, 0) \]
3. Vertex
The vertex of a parabola given by \( y = ax^2 + bx + c \) can also be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \]
Now, substitute \( x = 1 \) back into the function to find the y-coordinate of the vertex:
\[ y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121 \]
So, the vertex is:
\[ \text{vertex: } (1, -121) \]
Summary of Key Points
- X-Intercepts: \((-10, 0)\) and \((12, 0)\)
- Y-Intercept: \((0, -120)\)
- Vertex: \((1, -121)\)
Final answer:
- x-intercepts: \((-10, 0)\) and \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)