Question

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

(smaller) x-intercept: (
)

(larger) x-intercept: (
)

y-intercept: (
)

Answer 1

To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 48 \), we can complete the square or use the vertex formula and find the intercepts.

### Step 1: Find the Vertex
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

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

Here, \( a = 1 \) and \( b = -16 \):

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

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

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

Thus, the vertex is:

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

### Step 2: Find the x-intercepts
The x-intercepts occur where \( y = 0 \). Set the equation to zero and solve for \( x \):

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

We can factor the quadratic:

\[
0 = (x - 6)(x - 8)
\]

Setting each factor to zero gives the x-intercepts:

\[
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]
\[
x - 8 = 0 \quad \Rightarrow \quad x = 8
\]

The x-intercepts are at:

\[
\text{(smaller) x-intercept: } (6, 0)
\]
\[
\text{(larger) x-intercept: } (8, 0)
\]

### Step 3: Find the y-intercept
The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the original equation:

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

Thus, the y-intercept is:

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

### Summary of Key Points
- **Vertex:** (8, -16)
- **(smaller) x-intercept:** (6, 0)
- **(larger) x-intercept:** (8, 0)
- **y-intercept:** (0, 48)