Answer :
To solve the problem of identifying the vertex, intercepts, axis of symmetry, and then graphing the function [tex]\( f(x) = -x^2 - 6x - 11 \)[/tex], we will follow these steps:
### 1. Identify the Vertex
For a quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -11 \)[/tex].
First, let's determine the [tex]\( x \)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{-6}{2(-1)} = -\frac{6}{-2} = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-coordinate of the vertex is [tex]\( 3 \)[/tex].
Next, we substitute [tex]\( x = 3 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f(3) = -3^2 - 6(3) - 11 = -9 - 18 - 11 = -38 \][/tex]
Thus, the [tex]\( y \)[/tex]-coordinate of the vertex is [tex]\( -38 \)[/tex].
Therefore, the vertex is:
[tex]\[ \text{Vertex: } (3, -38) \][/tex]
### 2. Identify the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts are the points where [tex]\( f(x) = 0 \)[/tex]. To find the [tex]\( x \)[/tex]-intercepts, we solve the quadratic equation [tex]\( -x^2 - 6x - 11 = 0 \)[/tex].
The solutions to this equation are given by the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -11 \)[/tex]:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-11)}}{2(-1)} = \frac{6 \pm \sqrt{36 - 44}}{-2} = \frac{6 \pm \sqrt{-8}}{-2} \][/tex]
Since the discriminant [tex]\( 36 - 44 \)[/tex] is negative, the solutions involve imaginary numbers:
[tex]\[ x = \frac{6 \pm 2i\sqrt{2}}{-2} = -3 \pm i\sqrt{2} \][/tex]
This gives us the following [tex]\( x \)[/tex]-intercepts:
[tex]\[ \text{\( x \)-Intercepts: } (-3 - i\sqrt{2}, 0) \text{ and } (-3 + i\sqrt{2}, 0) \][/tex]
### 3. Identify the [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is the point where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) = -x^2 - 6x - 11 \)[/tex]:
[tex]\[ f(0) = -0^2 - 6(0) - 11 = -11 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ \text{\( y \)-Intercept: } (0, -11) \][/tex]
### 4. Identify the Axis of Symmetry
The axis of symmetry for the quadratic function is a vertical line that passes through the [tex]\( x \)[/tex]-coordinate of the vertex. Hence, the axis of symmetry is:
[tex]\[ \text{Axis of Symmetry: } x = 3 \][/tex]
### 5. Graph the Function
To graph the function [tex]\( f(x) = -x^2 - 6x - 11 \)[/tex], we plot the vertex, the intercepts, and draw the parabola:
1. Vertex: [tex]\((3, -38)\)[/tex]
2. [tex]\( x \)[/tex]-Intercepts: [tex]\((-3 - i\sqrt{2}, 0)\)[/tex] and [tex]\((-3 + i\sqrt{2}, 0)\)[/tex] (Note: These are complex and thus won't be visible on a real-number graph)
3. [tex]\( y \)[/tex]-Intercept: [tex]\((0, -11)\)[/tex]
4. Axis of Symmetry: The vertical line [tex]\( x = 3 \)[/tex]
Since the parabola opens downwards (because the coefficient of [tex]\( x^2 \)[/tex] is negative), it will be symmetrical around the line [tex]\( x = 3 \)[/tex]. The graph will pass through the points identified, curving downwards from the vertex.
Let's illustrate the graph below, considering the major points:
[tex]\[ \mathbf{Graph\ of\ } f(x) = -x^2 - 6x - 11 \][/tex]
(Unfortunately, I can't directly draw a graph here, but you can follow these points and plot on a coordinate plane or use graphing software.)
### 1. Identify the Vertex
For a quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -11 \)[/tex].
First, let's determine the [tex]\( x \)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{-6}{2(-1)} = -\frac{6}{-2} = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-coordinate of the vertex is [tex]\( 3 \)[/tex].
Next, we substitute [tex]\( x = 3 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f(3) = -3^2 - 6(3) - 11 = -9 - 18 - 11 = -38 \][/tex]
Thus, the [tex]\( y \)[/tex]-coordinate of the vertex is [tex]\( -38 \)[/tex].
Therefore, the vertex is:
[tex]\[ \text{Vertex: } (3, -38) \][/tex]
### 2. Identify the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts are the points where [tex]\( f(x) = 0 \)[/tex]. To find the [tex]\( x \)[/tex]-intercepts, we solve the quadratic equation [tex]\( -x^2 - 6x - 11 = 0 \)[/tex].
The solutions to this equation are given by the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -11 \)[/tex]:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-11)}}{2(-1)} = \frac{6 \pm \sqrt{36 - 44}}{-2} = \frac{6 \pm \sqrt{-8}}{-2} \][/tex]
Since the discriminant [tex]\( 36 - 44 \)[/tex] is negative, the solutions involve imaginary numbers:
[tex]\[ x = \frac{6 \pm 2i\sqrt{2}}{-2} = -3 \pm i\sqrt{2} \][/tex]
This gives us the following [tex]\( x \)[/tex]-intercepts:
[tex]\[ \text{\( x \)-Intercepts: } (-3 - i\sqrt{2}, 0) \text{ and } (-3 + i\sqrt{2}, 0) \][/tex]
### 3. Identify the [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is the point where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) = -x^2 - 6x - 11 \)[/tex]:
[tex]\[ f(0) = -0^2 - 6(0) - 11 = -11 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ \text{\( y \)-Intercept: } (0, -11) \][/tex]
### 4. Identify the Axis of Symmetry
The axis of symmetry for the quadratic function is a vertical line that passes through the [tex]\( x \)[/tex]-coordinate of the vertex. Hence, the axis of symmetry is:
[tex]\[ \text{Axis of Symmetry: } x = 3 \][/tex]
### 5. Graph the Function
To graph the function [tex]\( f(x) = -x^2 - 6x - 11 \)[/tex], we plot the vertex, the intercepts, and draw the parabola:
1. Vertex: [tex]\((3, -38)\)[/tex]
2. [tex]\( x \)[/tex]-Intercepts: [tex]\((-3 - i\sqrt{2}, 0)\)[/tex] and [tex]\((-3 + i\sqrt{2}, 0)\)[/tex] (Note: These are complex and thus won't be visible on a real-number graph)
3. [tex]\( y \)[/tex]-Intercept: [tex]\((0, -11)\)[/tex]
4. Axis of Symmetry: The vertical line [tex]\( x = 3 \)[/tex]
Since the parabola opens downwards (because the coefficient of [tex]\( x^2 \)[/tex] is negative), it will be symmetrical around the line [tex]\( x = 3 \)[/tex]. The graph will pass through the points identified, curving downwards from the vertex.
Let's illustrate the graph below, considering the major points:
[tex]\[ \mathbf{Graph\ of\ } f(x) = -x^2 - 6x - 11 \][/tex]
(Unfortunately, I can't directly draw a graph here, but you can follow these points and plot on a coordinate plane or use graphing software.)
