Question

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

Answer 1

To find the vertex and y-intercept of the quadratic function \( y = x^2 - 16x + 28 \), we can follow these steps:

### Vertex
The vertex of a quadratic function in the 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 given equation \( y = x^2 - 16x + 28 \):
- \( a = 1 \)
- \( b = -16 \)

Substituting into the formula gives:

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

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

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

Thus, the vertex of the quadratic function is:

\[
(8, -36)
\]

### Y-Intercept
The y-intercept of a function occurs when \( x = 0 \). We can calculate the y-intercept by substituting \( x = 0 \) into the equation:

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

Therefore, the y-intercept is:

\[
(0, 28)
\]

### Summary
- Vertex: \( (8, -36) \)
- Y-Intercept: \( (0, 28) \)