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