To analyze the quadratic function \( y = x^2 - 2x - 15 \), we will determine its x-intercepts, y-intercept, and vertex.
1. X-Intercepts
To find the x-intercepts, set \( y = 0 \) and solve for \( x \):
\[ 0 = x^2 - 2x - 15 \]
Factoring the quadratic:
\[ 0 = (x - 5)(x + 3) \]
Setting each factor to zero:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
The x-intercepts are:
- \( x = 5 \)
- \( x = -3 \)
2. Y-Intercept
To find the y-intercept, set \( x = 0 \):
\[ y = 0^2 - 2(0) - 15 = -15 \]
The y-intercept is:
- \( y = -15 \)
3. Vertex
The vertex of a parabola represented by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} \]
In our function \( y = x^2 - 2x - 15 \):
- \( a = 1 \)
- \( b = -2 \)
Calculating the x-coordinate of the vertex:
\[ x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1 \]
Now, substitute \( x = 1 \) back into the equation to find the y-coordinate of the vertex:
\[ y = (1)^2 - 2(1) - 15 = 1 - 2 - 15 = -16 \]
Thus, the vertex is:
- \( (1, -16) \)
Summary of Key Points
- X-Intercepts: \( (5, 0) \) and \( (-3, 0) \)
- Y-Intercept: \( (0, -15) \)
- Vertex: \( (1, -16) \)