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Magazine Chapter 9: Problem 265
Graph the function by using its properties. $$f(x)=-x^{2}-4 x+2$$
Short Answer
Expert verified
The vertex is at (2, -10), the parabola opens downward, y-intercept is (0, 2), x-intercepts are \( -2 \pm \sqrt{6} \).
Step by step solution
01
Identify the function and its form
Recognize that the function is a quadratic function in the standard form: \[ f(x) = -x^2 - 4x + 2 \] The general form of a quadratic function is \( f(x) = ax^2 + bx + c \), where \( a = -1 \), \( b = -4 \), and \( c = 2 \).
02
Determine the vertex
Use the vertex formula \( x = -\frac{b}{2a} \). For \( a = -1 \) and \( b = -4 \), this becomes: \[ x = -\frac{-4}{2(-1)} = 2 \]. Plug \( x = 2 \) back into the function to find \( y \): \[ f(2) = -(2)^2 - 4(2) + 2 = -4 - 8 + 2 = -10 \]. Thus, the vertex is at \( (2, -10) \).
03
Determine the axis of symmetry
The axis of symmetry of the parabola is a vertical line that passes through the vertex. For this function, it is given by \( x = 2 \).
04
Determine the direction of the parabola
Since \( a = -1 \), the coefficient of \( x^2 \) is negative. This means the parabola opens downward.
05
Find the y-intercept
Set \( x = 0 \) and calculate \( f(0) \): \[ f(0) = -0^2 - 4(0) + 2 = 2 \]. So, the y-intercept is \( (0, 2) \).
06
Find the x-intercepts
Solve the quadratic equation \( -x^2 - 4x + 2 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting in the values, we get: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(2)}}{2(-1)} \] \[ x = \frac{4 \pm \sqrt{16 + 8}}{-2} \] \[ x = \frac{4 \pm \sqrt{24}}{-2} \] \[ x = \frac{4 \pm 2\sqrt{6}}{-2} \] \[ x = -2 \pm \sqrt{6} \]. So, the x-intercepts are \( -2 + \sqrt{6} \) and \( -2 - \sqrt{6} \).
07
Plot the points and graph the function
Plot the vertex \( (2, -10) \), the y-intercept \( (0, 2) \), and the x-intercepts \( -2 + \sqrt{6} \) and \( -2 - \sqrt{6} \). Then, sketch the parabola opening downward through these points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A quadratic function is a type of polynomial that can be written in the standard form: \( f(x) = ax^2 + bx + c \).
Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.
Quadratic functions always produce a parabolic graph, which means they have a distinct 'U' shape.
In the provided exercise, the quadratic function is \( f(x) = -x^2 - 4x + 2 \).
From this, we can identify:
Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.
Quadratic functions always produce a parabolic graph, which means they have a distinct 'U' shape.
In the provided exercise, the quadratic function is \( f(x) = -x^2 - 4x + 2 \).
From this, we can identify:
- \( a = -1 \)
- \( b = -4 \)
- \( c = 2 \)
Vertex Formula
The vertex of a parabola represents its highest or lowest point, depending on its direction.
For a quadratic function in standard form, the x-coordinate of the vertex is found using the vertex formula: \( x = -\frac{b}{2a} \).
In the given function, \( a = -1 \) and \( b = -4 \). So, we calculate:
\( x = -\frac{-4}{2(-1)} = 2 \). We plug \( x = 2 \) back into the function to find the y-coordinate:
\[ f(2) = -(2)^2 - 4(2) + 2 = -4 - 8 + 2 = -10 \].
Thus, the vertex of the function is at \( (2, -10) \). The vertex helps us determine the maximum or minimum value of the function.
For a quadratic function in standard form, the x-coordinate of the vertex is found using the vertex formula: \( x = -\frac{b}{2a} \).
In the given function, \( a = -1 \) and \( b = -4 \). So, we calculate:
\( x = -\frac{-4}{2(-1)} = 2 \). We plug \( x = 2 \) back into the function to find the y-coordinate:
\[ f(2) = -(2)^2 - 4(2) + 2 = -4 - 8 + 2 = -10 \].
Thus, the vertex of the function is at \( (2, -10) \). The vertex helps us determine the maximum or minimum value of the function.
Axis of Symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves.
It passes through the vertex of the parabola and has the equation \( x = h \), where \( h \) is the x-coordinate of the vertex.
For the function \( f(x) = -x^2 - 4x + 2 \), we've found that the vertex is \( (2, -10) \).
Therefore, the axis of symmetry is the line \( x = 2 \).
This line helps us understand how the parabola is oriented in the coordinate plane.
It passes through the vertex of the parabola and has the equation \( x = h \), where \( h \) is the x-coordinate of the vertex.
For the function \( f(x) = -x^2 - 4x + 2 \), we've found that the vertex is \( (2, -10) \).
Therefore, the axis of symmetry is the line \( x = 2 \).
This line helps us understand how the parabola is oriented in the coordinate plane.
Direction of Parabola
The direction in which a parabola opens is determined by the coefficient \( a \).
This means the parabola opens downward. The direction of the parabola influences various attributes, like the maximum or minimum value and the overall appearance of the graph.
- If \( a > 0 \), the parabola opens upward (U-shaped).
- If \( a < 0 \), the parabola opens downward (n-shaped).
This means the parabola opens downward. The direction of the parabola influences various attributes, like the maximum or minimum value and the overall appearance of the graph.
Y-intercept
The y-intercept of a function is the point where the graph crosses the y-axis.
This happens when \( x = 0 \).
To find it, we substitute \( x = 0 \) into the function:
\[ f(0) = -0^2 - 4(0) + 2 = 2 \].
So, the y-intercept is \( (0, 2) \).
It's a starting point that helps in sketching the graph and provides a concrete reference for the function's behavior.
This happens when \( x = 0 \).
To find it, we substitute \( x = 0 \) into the function:
\[ f(0) = -0^2 - 4(0) + 2 = 2 \].
So, the y-intercept is \( (0, 2) \).
It's a starting point that helps in sketching the graph and provides a concrete reference for the function's behavior.
X-intercepts
The x-intercepts, also known as the roots or zeros of the function, are the points where the graph crosses the x-axis.
To find the x-intercepts, we solve the quadratic equation \( -x^2 - 4x + 2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
In the provided exercise, we have \( a = -1 \), \( b = -4 \), and \( c = 2 \).
Plugging in these values, we get: \[ x = \frac{4 \pm \sqrt{16 + 8}}{-2} = \frac{4 \pm \sqrt{24}}{-2} = \frac{4 \pm 2\sqrt{6}}{-2} = -2 \pm \sqrt{6} \].
Thus, the x-intercepts are \( -2 + \sqrt{6} \) and \( -2 - \sqrt{6} \).
X-intercepts are essential for plotting and understanding where the quadratic function changes direction.
To find the x-intercepts, we solve the quadratic equation \( -x^2 - 4x + 2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
In the provided exercise, we have \( a = -1 \), \( b = -4 \), and \( c = 2 \).
Plugging in these values, we get: \[ x = \frac{4 \pm \sqrt{16 + 8}}{-2} = \frac{4 \pm \sqrt{24}}{-2} = \frac{4 \pm 2\sqrt{6}}{-2} = -2 \pm \sqrt{6} \].
Thus, the x-intercepts are \( -2 + \sqrt{6} \) and \( -2 - \sqrt{6} \).
X-intercepts are essential for plotting and understanding where the quadratic function changes direction.
