Question
Consider the function f left parenthesis x right parenthesis equals 4 left parenthesis x minus 2 right parenthesis squared minus 5f(x)=4(x−2)2−5.
(a) Give the coordinates of the vertex of the graph of the function.
(b) Graph the function on a window that includes the vertex.
Question content area bottom
Part 1
(a) The vertex of the function is enter your response here. (Type an ordered pair.)
Part 2
(b) Choose the correct graph below.
A.
x y graph
B.
x y graph
C.
x y graph
D.
x y graph
[minus−10,10] by [minus−10,10], Xsclequals=1, Ysclequals=1
(a) Give the coordinates of the vertex of the graph of the function.
(b) Graph the function on a window that includes the vertex.
Question content area bottom
Part 1
(a) The vertex of the function is enter your response here. (Type an ordered pair.)
Part 2
(b) Choose the correct graph below.
A.
x y graph
B.
x y graph
C.
x y graph
D.
x y graph
[minus−10,10] by [minus−10,10], Xsclequals=1, Ysclequals=1
Answers
Answered by
GPT-4o mini
To analyze the function \( f(x) = 4(x - 2)^2 - 5 \), we can determine several key features:
### Part 1: Finding the Vertex
In the given function, \( f(x) = a(x - h)^2 + k \), the vertex (h, k) can be easily identified since the function is in vertex form.
1. **Identify the values for \( h \) and \( k \)**:
- \( h = 2 \) (the value of \( x \) that makes the expression \( (x - 2) \) equal to zero)
- \( k = -5 \) (the value of \( f(h) \) when \( x = 2 \))
2. **Therefore, the vertex is**:
\[
(h, k) = (2, -5)
\]
### Answer for Part 1:
The vertex of the function is \( \boxed{(2, -5)} \).
### Part 2: Graphing the Function
To graph this function, we can sketch it based on the vertex and other points.
1. **Vertex**:
The vertex is at \( (2, -5) \).
2. **Opening Direction**:
Since the coefficient of \( (x - 2)^2 \) (which is 4) is positive, the parabola opens upwards.
3. **Y-intercept**:
To find the y-intercept, set \( x = 0 \):
\[
f(0) = 4(0 - 2)^2 - 5 = 4(4) - 5 = 16 - 5 = 11
\]
Thus, the y-intercept is \( (0, 11) \).
4. **Additional Points**:
You can also find other points for more accuracy:
- For \( x = 1 \):
\[
f(1) = 4(1 - 2)^2 - 5 = 4(1) - 5 = 4 - 5 = -1 \quad \text{(point (1, -1))}
\]
- For \( x = 3 \):
\[
f(3) = 4(3 - 2)^2 - 5 = 4(1) - 5 = 4 - 5 = -1 \quad \text{(point (3, -1))}
\]
- For \( x = 4 \):
\[
f(4) = 4(4 - 2)^2 - 5 = 4(4) - 5 = 16 - 5 = 11 \quad \text{(point (4, 11))}
\]
### Choosing the Correct Graph
Given the vertex and points calculated, look for a graph that:
- Has a vertex at (2, -5).
- Opens upwards.
- Passes through the points calculated, including (0, 11), (1, -1), (3, -1), and (4, 11).
### Conclusion
You will need to look at the options provided to find the one that matches this description, which includes the vertex at (2, -5) and is consistent with the shape of a parabola that opens upwards and passes through the points mentioned.
### Part 1: Finding the Vertex
In the given function, \( f(x) = a(x - h)^2 + k \), the vertex (h, k) can be easily identified since the function is in vertex form.
1. **Identify the values for \( h \) and \( k \)**:
- \( h = 2 \) (the value of \( x \) that makes the expression \( (x - 2) \) equal to zero)
- \( k = -5 \) (the value of \( f(h) \) when \( x = 2 \))
2. **Therefore, the vertex is**:
\[
(h, k) = (2, -5)
\]
### Answer for Part 1:
The vertex of the function is \( \boxed{(2, -5)} \).
### Part 2: Graphing the Function
To graph this function, we can sketch it based on the vertex and other points.
1. **Vertex**:
The vertex is at \( (2, -5) \).
2. **Opening Direction**:
Since the coefficient of \( (x - 2)^2 \) (which is 4) is positive, the parabola opens upwards.
3. **Y-intercept**:
To find the y-intercept, set \( x = 0 \):
\[
f(0) = 4(0 - 2)^2 - 5 = 4(4) - 5 = 16 - 5 = 11
\]
Thus, the y-intercept is \( (0, 11) \).
4. **Additional Points**:
You can also find other points for more accuracy:
- For \( x = 1 \):
\[
f(1) = 4(1 - 2)^2 - 5 = 4(1) - 5 = 4 - 5 = -1 \quad \text{(point (1, -1))}
\]
- For \( x = 3 \):
\[
f(3) = 4(3 - 2)^2 - 5 = 4(1) - 5 = 4 - 5 = -1 \quad \text{(point (3, -1))}
\]
- For \( x = 4 \):
\[
f(4) = 4(4 - 2)^2 - 5 = 4(4) - 5 = 16 - 5 = 11 \quad \text{(point (4, 11))}
\]
### Choosing the Correct Graph
Given the vertex and points calculated, look for a graph that:
- Has a vertex at (2, -5).
- Opens upwards.
- Passes through the points calculated, including (0, 11), (1, -1), (3, -1), and (4, 11).
### Conclusion
You will need to look at the options provided to find the one that matches this description, which includes the vertex at (2, -5) and is consistent with the shape of a parabola that opens upwards and passes through the points mentioned.