Question
Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the domain and range of the function.
f left parenthesis x right parenthesis equals 2 x minus x squared minus 10
Question content area bottom left
Part 1
Use the graphing tool to graph the equation. Use the vertex and one of the intercepts when drawing the graph.
.
.
.
Question content area right
Part 1
-10
-8
-6
-4
-2
2
4
6
8
10
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
2
4
x
y
f left parenthesis x right parenthesis equals 2 x minus x squared minus 10
Question content area bottom left
Part 1
Use the graphing tool to graph the equation. Use the vertex and one of the intercepts when drawing the graph.
.
.
.
Question content area right
Part 1
-10
-8
-6
-4
-2
2
4
6
8
10
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
2
4
x
y
Answers
GPT-4o mini
To analyze the quadratic function \( f(x) = 2x - x^2 - 10 \) and sketch its graph, we follow these steps:
### Step 1: Rewrite the Function
First, let's rewrite the function in standard form. The given function can be rearranged as:
\[ f(x) = -x^2 + 2x - 10 \]
### Step 2: Find the Vertex
The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
In this case:
- \( a = -1 \)
- \( b = 2 \)
Calculating the x-coordinate of the vertex:
\[
x = -\frac{2}{2(-1)} = 1
\]
Now, substitute \( x = 1 \) into the function to find the corresponding y-coordinate:
\[
f(1) = -1^2 + 2(1) - 10 = -1 + 2 - 10 = -9
\]
So, the vertex is at the point \( (1, -9) \).
### Step 3: Find the Intercepts
#### y-intercept:
To find the y-intercept, set \( x = 0 \):
\[
f(0) = 2(0) - (0)^2 - 10 = -10
\]
Thus, the y-intercept is at the point \( (0, -10) \).
#### x-intercepts:
To find the x-intercepts, set \( f(x) = 0 \):
\[
0 = -x^2 + 2x - 10
\]
Rearranging gives us:
\[
x^2 - 2x + 10 = 0
\]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 10 \):
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)} = \frac{2 \pm \sqrt{4 - 40}}{2}
\]
\[
= \frac{2 \pm \sqrt{-36}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i
\]
Since we get complex roots, there are no x-intercepts on the real plane, indicating the parabola does not cross the x-axis.
### Step 4: Axis of Symmetry
The axis of symmetry of the parabola is a vertical line that passes through the vertex. Its equation is:
\[
x = 1
\]
### Step 5: Domain and Range
- **Domain**: The domain of a quadratic function is all real numbers: \( (-\infty, +\infty) \).
- **Range**: Since the coefficient of \( x^2 \) is negative (indicating the parabola opens downward) and the vertex is at \( (1, -9) \), the range is all values \( y \) that are less than or equal to -9:
\[
(-\infty, -9]
\]
### Sketching the Graph
Now let’s summarize the key points for the sketch:
- Vertex: \( (1, -9) \)
- y-intercept: \( (0, -10) \)
- No x-intercepts
- Axis of symmetry: \( x = 1 \)
#### Graph
Plot the vertex and y-intercept on a graph and sketch the parabola opening downwards. Make sure it is symmetric about the line \( x = 1 \).
### Conclusion
The domain of the function is \( (-\infty, +\infty) \) and the range is \( (-\infty, -9] \).
### Step 1: Rewrite the Function
First, let's rewrite the function in standard form. The given function can be rearranged as:
\[ f(x) = -x^2 + 2x - 10 \]
### Step 2: Find the Vertex
The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
In this case:
- \( a = -1 \)
- \( b = 2 \)
Calculating the x-coordinate of the vertex:
\[
x = -\frac{2}{2(-1)} = 1
\]
Now, substitute \( x = 1 \) into the function to find the corresponding y-coordinate:
\[
f(1) = -1^2 + 2(1) - 10 = -1 + 2 - 10 = -9
\]
So, the vertex is at the point \( (1, -9) \).
### Step 3: Find the Intercepts
#### y-intercept:
To find the y-intercept, set \( x = 0 \):
\[
f(0) = 2(0) - (0)^2 - 10 = -10
\]
Thus, the y-intercept is at the point \( (0, -10) \).
#### x-intercepts:
To find the x-intercepts, set \( f(x) = 0 \):
\[
0 = -x^2 + 2x - 10
\]
Rearranging gives us:
\[
x^2 - 2x + 10 = 0
\]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 10 \):
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)} = \frac{2 \pm \sqrt{4 - 40}}{2}
\]
\[
= \frac{2 \pm \sqrt{-36}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i
\]
Since we get complex roots, there are no x-intercepts on the real plane, indicating the parabola does not cross the x-axis.
### Step 4: Axis of Symmetry
The axis of symmetry of the parabola is a vertical line that passes through the vertex. Its equation is:
\[
x = 1
\]
### Step 5: Domain and Range
- **Domain**: The domain of a quadratic function is all real numbers: \( (-\infty, +\infty) \).
- **Range**: Since the coefficient of \( x^2 \) is negative (indicating the parabola opens downward) and the vertex is at \( (1, -9) \), the range is all values \( y \) that are less than or equal to -9:
\[
(-\infty, -9]
\]
### Sketching the Graph
Now let’s summarize the key points for the sketch:
- Vertex: \( (1, -9) \)
- y-intercept: \( (0, -10) \)
- No x-intercepts
- Axis of symmetry: \( x = 1 \)
#### Graph
Plot the vertex and y-intercept on a graph and sketch the parabola opening downwards. Make sure it is symmetric about the line \( x = 1 \).
### Conclusion
The domain of the function is \( (-\infty, +\infty) \) and the range is \( (-\infty, -9] \).