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 4 x minus x squared minus 5
f left parenthesis x right parenthesis equals 4 x minus x squared minus 5
Answers
GPT-4o mini
To sketch the graph of the quadratic function \( f(x) = 4x - x^2 - 5 \), we'll first rewrite it in a more standard form and then find the vertex, the intercepts, the axis of symmetry, and the domain and range.
### Step 1: Rewriting the function
Let's rewrite the function in standard form:
\[
f(x) = -x^2 + 4x - 5
\]
This is a downward-opening parabola.
### Step 2: Finding the vertex
The vertex of a quadratic function in standard form \( ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
In our function:
- \( a = -1 \)
- \( b = 4 \)
Calculating the x-coordinate of the vertex:
\[
x = -\frac{4}{2 \cdot -1} = 2
\]
Now, substitute \( x = 2 \) back into the function to find the y-coordinate of the vertex:
\[
f(2) = -2^2 + 4(2) - 5 = -4 + 8 - 5 = -1
\]
So, the vertex is at \( (2, -1) \).
### Step 3: Finding the x-intercepts
To find the x-intercepts, we set \( f(x) = 0 \):
\[
-x^2 + 4x - 5 = 0
\]
Rearranging gives:
\[
x^2 - 4x + 5 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2}
\]
Since the discriminant \( (16 - 20) < 0 \), there are no real x-intercepts (the parabola does not intersect the x-axis).
### Step 4: Finding the y-intercept
To find the y-intercept, we evaluate \( f(0) \):
\[
f(0) = -0^2 + 4(0) - 5 = -5
\]
So, the y-intercept is \( (0, -5) \).
### Step 5: Axis of symmetry
The axis of symmetry for the parabola can be described by the line:
\[
x = 2
\]
### Step 6: Domain and range
- **Domain**: The domain of a quadratic function is all real numbers, so:
\[
\text{Domain: } (-\infty, \infty)
\]
- **Range**: Since the parabola opens downward and the vertex is the maximum point, the range will be all y-values less than or equal to the y-coordinate of the vertex:
\[
\text{Range: } (-\infty, -1]
\]
### Final Summary
- **Vertex**: \( (2, -1) \)
- **Y-intercept**: \( (0, -5) \)
- **X-intercepts**: None
- **Axis of symmetry**: \( x = 2 \)
- **Domain**: \( (-\infty, \infty) \)
- **Range**: \( (-\infty, -1] \)
### Sketch
1. Draw coordinate axes.
2. Plot the vertex at \( (2, -1) \).
3. Plot the y-intercept at \( (0, -5) \).
4. Since there are no x-intercepts, note that the parabola opens downward starting at the vertex and extending to the left and right.
5. Notice that the parabola never crosses the x-axis.
The graph will look like a downward-opening parabola with its vertex at \( (2, -1) \), extending downwards below the x-axis without intersecting it.
### Step 1: Rewriting the function
Let's rewrite the function in standard form:
\[
f(x) = -x^2 + 4x - 5
\]
This is a downward-opening parabola.
### Step 2: Finding the vertex
The vertex of a quadratic function in standard form \( ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
In our function:
- \( a = -1 \)
- \( b = 4 \)
Calculating the x-coordinate of the vertex:
\[
x = -\frac{4}{2 \cdot -1} = 2
\]
Now, substitute \( x = 2 \) back into the function to find the y-coordinate of the vertex:
\[
f(2) = -2^2 + 4(2) - 5 = -4 + 8 - 5 = -1
\]
So, the vertex is at \( (2, -1) \).
### Step 3: Finding the x-intercepts
To find the x-intercepts, we set \( f(x) = 0 \):
\[
-x^2 + 4x - 5 = 0
\]
Rearranging gives:
\[
x^2 - 4x + 5 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2}
\]
Since the discriminant \( (16 - 20) < 0 \), there are no real x-intercepts (the parabola does not intersect the x-axis).
### Step 4: Finding the y-intercept
To find the y-intercept, we evaluate \( f(0) \):
\[
f(0) = -0^2 + 4(0) - 5 = -5
\]
So, the y-intercept is \( (0, -5) \).
### Step 5: Axis of symmetry
The axis of symmetry for the parabola can be described by the line:
\[
x = 2
\]
### Step 6: Domain and range
- **Domain**: The domain of a quadratic function is all real numbers, so:
\[
\text{Domain: } (-\infty, \infty)
\]
- **Range**: Since the parabola opens downward and the vertex is the maximum point, the range will be all y-values less than or equal to the y-coordinate of the vertex:
\[
\text{Range: } (-\infty, -1]
\]
### Final Summary
- **Vertex**: \( (2, -1) \)
- **Y-intercept**: \( (0, -5) \)
- **X-intercepts**: None
- **Axis of symmetry**: \( x = 2 \)
- **Domain**: \( (-\infty, \infty) \)
- **Range**: \( (-\infty, -1] \)
### Sketch
1. Draw coordinate axes.
2. Plot the vertex at \( (2, -1) \).
3. Plot the y-intercept at \( (0, -5) \).
4. Since there are no x-intercepts, note that the parabola opens downward starting at the vertex and extending to the left and right.
5. Notice that the parabola never crosses the x-axis.
The graph will look like a downward-opening parabola with its vertex at \( (2, -1) \), extending downwards below the x-axis without intersecting it.