Question

What are the key points on the graph of y = x^2 − 16x + 48? Name the vertex, x-intercepts, and y-intercept.
vertex: (___)
(smaller) x-intercepts: (___)
(larger) x-intercept: (___)
y-intercept: (___)

Answer 1

To analyze the quadratic function \( y = x^2 - 16x + 48 \), we will determine the vertex, x-intercepts, and y-intercept.

### 1. Finding the Vertex
The vertex of a quadratic \( y = ax^2 + bx + c \) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

For the function \( y = x^2 - 16x + 48 \):
- \( a = 1 \)
- \( b = -16 \)

Substituting into the formula:

\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]

To find the y-coordinate of the vertex, substitute \( x = 8 \) back into the equation:

\[
y = 8^2 - 16 \cdot 8 + 48
\]
\[
y = 64 - 128 + 48 = -16
\]

Thus, the vertex is at:

\[
\text{vertex: } (8, -16)
\]

### 2. Finding the x-intercepts
To find the x-intercepts, set \( y = 0 \):

\[
0 = x^2 - 16x + 48
\]

This is a standard form quadratic equation. We can factor it:

\[
0 = (x - 12)(x - 4)
\]

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:

\[
\text{smaller } x\text{-intercept: } (4, 0)
\]
\[
\text{larger } x\text{-intercept: } (12, 0)
\]

### 3. Finding the y-intercept
To find the y-intercept, set \( x = 0 \):

\[
y = 0^2 - 16 \cdot 0 + 48 = 48
\]

Thus, the y-intercept is:

\[
\text{y-intercept: } (0, 48)
\]

### Summary of Key Points
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)