To analyze the graph of the quadratic function \( y = x^2 - 2x - 15 \), we can find the x-intercepts, y-intercept, and vertex.
1. X-Intercepts:
To find the x-intercepts, we set \( y = 0 \):
\[ 0 = x^2 - 2x - 15 \]
To solve this quadratic equation, we can factor it:
\[ 0 = (x - 5)(x + 3) \]
Setting each factor equal to zero gives the x-intercepts:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
Thus, the x-intercepts are \( (5, 0) \) and \( (-3, 0) \).
2. Y-Intercept:
To find the y-intercept, we set \( x = 0 \):
\[ y = (0)^2 - 2(0) - 15 = -15 \]
Thus, the y-intercept is \( (0, -15) \).
3. Vertex:
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \( a = 1 \) and \( b = -2 \):
\[ x = -\frac{-2}{2(1)} = \frac{2}{2} = 1 \]
Now, we can find the corresponding \( y \)-coordinate by substituting \( x = 1 \) back into the original equation:
\[ y = (1)^2 - 2(1) - 15 = 1 - 2 - 15 = -16 \]
Thus, the vertex is at \( (1, -16) \).
Summary:
- X-Intercepts: \( (5, 0) \) and \( (-3, 0) \)
- Y-Intercept: \( (0, -15) \)
- Vertex: \( (1, -16) \)