Question

What are the key points on the graph of y = x2 − 16x + 28
?

(6 points)
Vertex: (
,
)

Smaller x-intercept: (
, 0)

Larger x-intercept (
, 0)

y- intercept: (0,
)

Answer 1

what's the axis of symmetry

Answer 2

To find the key points of the quadratic function \( y = x^2 - 16x + 28 \), we need to identify the vertex, x-intercepts, and y-intercept.

### Step 1: Vertex
The vertex of a parabola defined by \( y = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
where \( a = 1 \) and \( b = -16 \).

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

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

So, the vertex is:
\[
\text{Vertex: } (8, -36)
\]

### Step 2: x-intercepts
To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 16x + 28
\]

We will solve this quadratic using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -16 \), and \( c = 28 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 28 = 256 - 112 = 144
\]

Now, using the quadratic formula:
\[
x = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2}
\]

Calculating the two possible values for \( x \):
\[
x_1 = \frac{28}{2} = 14 \quad \text{(larger x-intercept)}
\]
\[
x_2 = \frac{4}{2} = 2 \quad \text{(smaller x-intercept)}
\]

Thus, the x-intercepts are:
\[
\text{Smaller x-intercept: } (2, 0)
\]
\[
\text{Larger x-intercept: } (14, 0)
\]

### Step 3: y-intercept
To find the y-intercept, substitute \( x = 0 \) into the equation:
\[
y = 0^2 - 16 \cdot 0 + 28 = 28
\]
So, the y-intercept is:
\[
\text{y-intercept: } (0, 28)
\]

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

### Final Answer
```
Vertex: (8, -36)

Smaller x-intercept: (2, 0)

Larger x-intercept: (14, 0)

y-intercept: (0, 28)
```

Answer 3

The axis of symmetry for a quadratic function in the form \( y = ax^2 + bx + c \) is given by the equation:

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

In the case of the function \( y = x^2 - 16x + 28 \), we have:

- \( a = 1 \)
- \( b = -16 \)

Plugging in these values:

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

Therefore, the axis of symmetry is:

\[
\text{Axis of symmetry: } x = 8
\]