Question

What are the key points on the graph of y=x^2−16x+28 ? 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 \( y = x^2 - 16x + 28 \), we can determine the vertex, x-intercepts, and y-intercept.

### 1. **Vertex:**
The vertex form of a quadratic equation is given by the formula \( h = -\frac{b}{2a} \) for the x-coordinate, where \( a \) is the coefficient of \( x^2 \) and \( b \) is the coefficient of \( x \).

For our equation:
- \( a = 1 \)
- \( b = -16 \)

Calculating \( h \):
\[
h = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]

Now substitute \( x = 8 \) back into the equation to find the y-coordinate:
\[
y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36
\]

Thus, the vertex is at \( (8, -36) \).

### 2. **X-Intercepts:**
To find the x-intercepts, set \( y = 0 \) and solve:
\[
0 = x^2 - 16x + 28
\]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(28)}}{2(1)} = \frac{16 \pm \sqrt{256 - 112}}{2}
\]
\[
= \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2}
\]

Calculating the two x-values:
1. \( x = \frac{28}{2} = 14 \)
2. \( x = \frac{4}{2} = 2 \)

Thus, the x-intercepts are \( (2, 0) \) and \( (14, 0) \).

### 3. **Y-Intercept:**
To find the y-intercept, set \( x = 0 \):
\[
y = (0)^2 - 16(0) + 28 = 28
\]

Thus, the y-intercept is \( (0, 28) \).

### Summary of Key Points:
- **Vertex:** \( (8, -36) \)
- **Smaller x-intercept:** \( (2, 0) \)
- **Larger x-intercept:** \( (14, 0) \)
- **Y-intercept:** \( (0, 28) \)

### Final Answers:
- vertex: **(8, -36)**
- (smaller) x-intercept: **(2, 0)**
- (larger) x-intercept: **(14, 0)**
- y-intercept: **(0, 28)**