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
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Answer 1

To find the key points of the quadratic function \( y = x^2 - 16x + 48 \), we will 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:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 1 \) and \( b = -16 \).

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

Now, substituting \( x = 8 \) back into the function to find the y-coordinate of the vertex:

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

So, the vertex is \( (8, -16) \).

2. **X-intercepts**: To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 16x + 48
\]
Factoring the quadratic:
\[
0 = (x - 12)(x - 4)
\]

Setting each factor equal to zero gives:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

Thus, the smaller x-intercept is \( (4, 0) \) and the larger x-intercept is \( (12, 0) \).

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

Putting it all together:

- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)

Final answer:
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)