Question

What are the key points on the graph of y=x2−16x+48
? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: (
)

(smaller) x-intercept: (
)

(larger) x-intercept: (
)

y-intercept: (
)

Remaining Attempts : 3

Answer 1

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

### 1. Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
For our function:
- \( a = 1 \)
- \( b = -16 \)
- \( c = 48 \)

Calculating the x-coordinate of the vertex:
\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]

Now we find the y-coordinate by substituting \( x = 8 \) back into the equation:
\[
y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]

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

### 2. X-intercepts
To find the x-intercepts, we set \( y = 0 \):
\[
0 = x^2 - 16x + 48
\]
We can solve this quadratic by factoring or using the quadratic formula. Let's first factor it:
\[
x^2 - 16x + 48 = (x - 12)(x - 4) = 0
\]
This gives us the x-intercepts:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12 \\
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

Thus, we have:
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)

### 3. Y-intercept
To find the y-intercept, we set \( x = 0 \):
\[
y = (0)^2 - 16(0) + 48 = 48
\]
So 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: \( (0, 48) \)

### Final Answers
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)