Question

What are the key points in the graph of y equals x squared - 16x + 48 name the vertex X intercepts and y-intercepts

Answer 1

To analyze the quadratic function \( y = x^2 - 16x + 48 \), let's identify the key points: the vertex, x-intercepts, and y-intercept.

1. **Vertex**: The vertex of a quadratic function \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
For the equation \( y = x^2 - 16x + 48 \), we have \( a = 1 \) and \( b = -16 \). Plugging in these values:
\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]
To find the y-coordinate, substitute \( x = 8 \) back into the equation:
\[
y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]
Thus, the vertex is \( (8, -16) \).

2. **Y-intercept**: The y-intercept occurs where \( x = 0 \):
\[
y = 0^2 - 16(0) + 48 = 48
\]
Thus, the y-intercept is \( (0, 48) \).

3. **X-intercepts**: The x-intercepts occur where \( y = 0 \). To find the x-intercepts, we solve the equation:
\[
0 = x^2 - 16x + 48
\]
We can factor this quadratic:
\[
0 = (x - 12)(x - 4)
\]
Setting each factor to zero gives:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
Thus, the x-intercepts are \( (12, 0) \) and \( (4, 0) \).

### Summary of Key Points:
- **Vertex**: \( (8, -16) \)
- **Y-intercept**: \( (0, 48) \)
- **X-intercepts**: \( (12, 0) \) and \( (4, 0) \)