Question

What are the key points on the graph of y=x2−16x+48y=x2−16x+48? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: () (smaller) x-intercept: () (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. Vertex
The vertex of a parabola defined by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).

Here, \( a = 1 \), \( b = -16 \), and \( c = 48 \).

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

Now we substitute \( x = 8 \) back into the equation to find the y-coordinate:

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

Thus, the vertex is:
\[
\text{Vertex: } (8, -16)
\]

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

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

We can factor this quadratic equation:

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

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:
\[
\text{Smaller x-intercept: } (4, 0), \quad \text{Larger x-intercept: } (12, 0)
\]

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

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

Thus, the y-intercept is:
\[
\text{Y-intercept: } (0, 48)
\]

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