Question

What are the key points on the graph of y=x2−2x−15
?(4 points)
x-intercepts:
and

y-intercept:

vertex:

Answer 1

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:** \( (1, -16) \)