Question

What are the key points on the graph of y=x2−16x+28? Name the vertex, x-intercepts, and y-intercept.(2 points)

Answer 1

To analyze the quadratic function \( y = x^2 - 16x + 28 \), we can determine key points such as the vertex, x-intercepts, and y-intercept.

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

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

For the equation \( y = x^2 - 16x + 28 \):
- \( a = 1 \)
- \( b = -16 \)

Calculating the x-coordinate of the vertex:

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

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

\[
y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36
\]

Thus, the vertex is at:

\[
\text{Vertex: } (8, -36)
\]

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

\[
0 = x^2 - 16x + 28
\]

We can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(28)}}{2(1)}
\]
\[
x = \frac{16 \pm \sqrt{256 - 112}}{2}
\]
\[
x = \frac{16 \pm \sqrt{144}}{2}
\]
\[
x = \frac{16 \pm 12}{2}
\]

Calculating the two potential x-values:

1. \( x = \frac{16 + 12}{2} = \frac{28}{2} = 14 \)
2. \( x = \frac{16 - 12}{2} = \frac{4}{2} = 2 \)

Thus, the x-intercepts are:

\[
\text{X-intercepts: } (14, 0) \text{ and } (2, 0)
\]

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

\[
y = (0)^2 - 16(0) + 28 = 28
\]

Thus, the y-intercept is at:

\[
\text{Y-intercept: } (0, 28)
\]

### Summary of Key Points:
- **Vertex:** \( (8, -36) \)
- **X-intercepts:** \( (14, 0) \) and \( (2, 0) \)
- **Y-intercept:** \( (0, 28) \)