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Magazine Chapter 2: Problem 20
In exercises \(17-20,\) complete the square and use your result to help you graph the function. $$ f(x)-x^{2}-4 x+6 $$
Short Answer
Expert verified
The function is \( f(x) = (x-2)^2 + 2 \) and its vertex is \((2, 2)\).
Step by step solution
01
Arrange the Equation
First, rewrite the function clearly as an expression of a quadratic function. The given function is \( f(x) = x^2 - 4x + 6 \). Notice it is already arranged in descending order of the power of \( x \).
02
Identify Parts for Completing the Square
Look at the quadratic term \( x^2 \) and the linear term \(-4x \). To complete the square, focus on \( x^2 - 4x \).
03
Completing the Square
To complete the square, we take the coefficient of \( x \) (which is \(-4\)), divide it by 2, and square the result. This procedure gives us: \(-4/2 = -2\), and \((-2)^2 = 4\). Add and subtract this square completing term inside the function: \( f(x) = (x^2 - 4x + 4) + 6 - 4 \). The expression within the parenthesis is a complete square.
04
Rewrite as a Perfect Square
The complete square expression \( x^2 - 4x + 4 \) can be rewritten as \( (x - 2)^2 \) using the identity \( a^2 - 2ab + b^2 = (a - b)^2 \). Thus, \( f(x) = (x - 2)^2 + 2 \).
05
Graphing the Function
The completed square form \( (x-2)^2 + 2 \) shows a vertex form of the function, where the vertex \( (h, k) \) of \( (x-h)^2 + k \) is at point \( (2, 2) \). Since the coefficient of \( (x-2)^2 \) is positive, the parabola opens upwards. Graph this function with the vertex at \( (2, 2) \) and axis of symmetry \( x = 2 \). The \( y \)-intercept from the original equation is at \( (0, 6) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a helpful algebraic technique often used to transform a quadratic expression into a form that makes it easier to solve or graph.
To start, identify the quadratic and the linear coefficient in the expression.
For instance, in the expression \( x^2 - 4x + 6 \), focus on \( x^2 - 4x \). The trick is to make this expression into a perfect square trinomial.Here's how to do it:
To start, identify the quadratic and the linear coefficient in the expression.
For instance, in the expression \( x^2 - 4x + 6 \), focus on \( x^2 - 4x \). The trick is to make this expression into a perfect square trinomial.Here's how to do it:
- Take the linear coefficient, which is \(-4\), divide it by 2 to get \(-2\), and then square it to get \(4\).
- Add and subtract this number inside the expression. Thus, \( x^2 - 4x \) becomes \((x^2 - 4x + 4) - 4\).
Quadratic Functions
Quadratic functions are polynomial expressions of degree 2.
These functions are in the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
Here, the function \( f(x) = x^2 - 4x + 6 \) is a quadratic function because it contains an \( x^2 \) term.Key features of quadratic functions include:
These functions are in the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
Here, the function \( f(x) = x^2 - 4x + 6 \) is a quadratic function because it contains an \( x^2 \) term.Key features of quadratic functions include:
- The parabolic shape of their graphs.
- The axis of symmetry, which runs vertically through the vertex.
- The vertex, which is the peak or trough depending on the direction of the parabola.
Vertex Form
The vertex form of a quadratic function makes identifying the vertex of the parabola simple, providing a direct view of the shape's most important feature.
The vertex form looks like this: \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex.
In our exercise, after completing the square, the function is rewritten as \( f(x) = (x-2)^2 + 2 \).This form immediately tells us:
The vertex form looks like this: \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex.
In our exercise, after completing the square, the function is rewritten as \( f(x) = (x-2)^2 + 2 \).This form immediately tells us:
- The vertex of the parabola is at \((h, k) = (2, 2)\).
- The parabola opens upwards because the coefficient of \((x-2)^2\) is positive.
Graphing Parabolas
Graphing parabolas involves plotting the quadratic equation and understanding its shape.
After converting the quadratic function into vertex form, graphing becomes more intuitive.
Using our example \( f(x) = (x-2)^2 + 2 \), notice where key points on the graph should be.Some steps to follow when graphing:
After converting the quadratic function into vertex form, graphing becomes more intuitive.
Using our example \( f(x) = (x-2)^2 + 2 \), notice where key points on the graph should be.Some steps to follow when graphing:
- Plot the vertex \((2, 2)\) as it is the highest or lowest point, depending on the parabola's orientation.
- Since it opens upwards, the arms of the parabola extend away from this vertex.
- Identify the axis of symmetry, which is the line \( x = 2 \).
- Plot additional points, such as the \( y \)-intercept \((0, 6)\), for reference.
