Question
16.
Use the image to answer the questions.
(5 points)
Identify these key features of the parabola.
vertex: (
,
)
Smaller x-intercept : (
, 0)
larger x -intercept: (
, 0)
y-intercept: (0,
)
Question 2
17.
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,
)
Axis of symmetry: x=
Skip to navigation
Use the image to answer the questions.
(5 points)
Identify these key features of the parabola.
vertex: (
,
)
Smaller x-intercept : (
, 0)
larger x -intercept: (
, 0)
y-intercept: (0,
)
Question 2
17.
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,
)
Axis of symmetry: x=
Skip to navigation
Answers
Answered by
GPT-4o mini
To find the key features of the parabola given by the equation \( y = x^2 - 16x + 28 \), we can follow these steps:
### Step 1: Find the vertex
The vertex of a parabola in the standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
For the equation \( y = x^2 - 16x + 28 \):
- \( a = 1 \)
- \( b = -16 \)
- \( c = 28 \)
**Calculate x-coordinate of the vertex:**
\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]
**Calculate y-coordinate of the vertex:**
\[
y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36
\]
**Vertex:**
\[
(8, -36)
\]
### Step 2: Find x-intercepts
To find the x-intercepts, we set \( y = 0 \):
\[
0 = x^2 - 16x + 28
\]
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Plugging \( a = 1, b = -16, c = 28 \):
\[
x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1}
\]
\[
x = \frac{16 \pm \sqrt{256 - 112}}{2}
\]
\[
x = \frac{16 \pm \sqrt{144}}{2}
\]
\[
x = \frac{16 \pm 12}{2}
\]
This gives us two solutions:
1. \( x = \frac{28}{2} = 14 \) (larger x-intercept)
2. \( x = \frac{4}{2} = 2 \) (smaller x-intercept)
**X-intercepts:**
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
### Step 3: Find y-intercept
To find the y-intercept, set \( x = 0 \):
\[
y = 0^2 - 16(0) + 28 = 28
\]
**Y-intercept:**
\[
(0, 28)
\]
### Step 4: Axis of symmetry
The axis of symmetry is the x-coordinate of the vertex:
\[
x = 8
\]
### Summary of the Key Features:
1. **Vertex:** \( (8, -36) \)
2. **Smaller x-intercept:** \( (2, 0) \)
3. **Larger x-intercept:** \( (14, 0) \)
4. **Y-intercept:** \( (0, 28) \)
5. **Axis of symmetry:** \( x = 8 \)
You can fill in the responses as specified in your question.
### Step 1: Find the vertex
The vertex of a parabola in the standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
For the equation \( y = x^2 - 16x + 28 \):
- \( a = 1 \)
- \( b = -16 \)
- \( c = 28 \)
**Calculate x-coordinate of the vertex:**
\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]
**Calculate y-coordinate of the vertex:**
\[
y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36
\]
**Vertex:**
\[
(8, -36)
\]
### Step 2: Find x-intercepts
To find the x-intercepts, we set \( y = 0 \):
\[
0 = x^2 - 16x + 28
\]
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Plugging \( a = 1, b = -16, c = 28 \):
\[
x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1}
\]
\[
x = \frac{16 \pm \sqrt{256 - 112}}{2}
\]
\[
x = \frac{16 \pm \sqrt{144}}{2}
\]
\[
x = \frac{16 \pm 12}{2}
\]
This gives us two solutions:
1. \( x = \frac{28}{2} = 14 \) (larger x-intercept)
2. \( x = \frac{4}{2} = 2 \) (smaller x-intercept)
**X-intercepts:**
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
### Step 3: Find y-intercept
To find the y-intercept, set \( x = 0 \):
\[
y = 0^2 - 16(0) + 28 = 28
\]
**Y-intercept:**
\[
(0, 28)
\]
### Step 4: Axis of symmetry
The axis of symmetry is the x-coordinate of the vertex:
\[
x = 8
\]
### Summary of the Key Features:
1. **Vertex:** \( (8, -36) \)
2. **Smaller x-intercept:** \( (2, 0) \)
3. **Larger x-intercept:** \( (14, 0) \)
4. **Y-intercept:** \( (0, 28) \)
5. **Axis of symmetry:** \( x = 8 \)
You can fill in the responses as specified in your question.