Problem 66
Question
Determine whether each function is written in vertex form. If a function is not in vertex form, rewrite the function. $$ y=0.5 x^{2}+10 $$
Step-by-Step Solution
Verified Answer
The function is not initially in vertex form, but it can be rewritten as \(y=0.5(x-0)^2+10\) which is in vertex form.
1Step 1: Identify Vertex Form of a Quadratic Function
The vertex form of a quadratic function is given by the equation \(y=a(x-h)^2+k\), where \((h,k)\) is the vertex of the parabola. Here, \(a\) determines the width and direction of the parabola.
2Step 2: Compare Given Function With Vertex Form
The provided quadratic function is \(y=0.5x^2+10\). This needs to be compared to the vertex form to determine if it's already in the correct format.
3Step 3: Determine if Function is in Vertex Form
Since the given function does not have an \((x-h)^2\) term, it is not in vertex form.
4Step 4: Rewrite Function in Vertex Form
The given function can be rewritten by completing the square if necessary. However, in this case, the given function is already a perfect square trinomial minus the constant term \(k\). So, the given equation can be written as \(y=0.5(x-0)^2+10\), which is in vertex form with the vertex at \((0,10)\).
Key Concepts
Quadratic FunctionsCompleting the SquareParabola Vertex
Quadratic Functions
Quadratic functions are fundamental in algebra and form parabolic shapes on a graph. They can be written in the standard form, which is \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). The coefficient \(a\) affects the parabola's width and concavity—whether it opens upwards (if \(a > 0\)) or downwards (if \(a < 0\)).
In the context of our problem, the quadratic function provided is \(y = 0.5x^2 + 10\), which indicates a parabola that opens upwards due to the positive coefficient \(0.5\) in front of \(x^2\). This function, however, is not presented in vertex form, which focuses on detailing the parabola's vertex.
In the context of our problem, the quadratic function provided is \(y = 0.5x^2 + 10\), which indicates a parabola that opens upwards due to the positive coefficient \(0.5\) in front of \(x^2\). This function, however, is not presented in vertex form, which focuses on detailing the parabola's vertex.
- Standard form: Focuses on the \(x\)- and \(y\)-intercepts of the graph.
- Vertex form: Focuses on the vertex or the highest/lowest point of the parabola.
Completing the Square
Completing the square is a technique used to manipulate an equation so that we can express it in its vertex form. This method involves creating a perfect square trinomial from a quadratic equation, which then can be written as \(a(x - h)^2 + k\).
To complete the square, we need to:
To complete the square, we need to:
- Divide the coefficient of \(x^2\) by 2.
- Square this result to find the 'completing term'.
- Add and subtract this completing term within the equation to maintain balance.
Parabola Vertex
The vertex of a parabola is a crucial point as it represents the maximum or minimum value of the quadratic function depending on whether the parabola opens downward or upward, respectively. The vertex form, \(y=a(x-h)^2+k\), explicitly shows the vertex through \(h\) and \(k\).
For a function in vertex form, the vertex \( (h, k) \) provides not only the exact coordinates of the turn or tip of the parabola but also offers the axis of symmetry of the graph, which is the line \(x = h\).
In our exercise, the initial function was not in vertex form, but after rewriting, we found that the vertex is at \( (0, 10) \). The axis of symmetry is the line \(x = 0\), i.e., the y-axis itself. This means our parabola is symmetrical about the y-axis and has its minimum point at \(y = 10\).Identifying the vertex helps in sketching the graph accurately and understanding the behavior of the quadratic function.
For a function in vertex form, the vertex \( (h, k) \) provides not only the exact coordinates of the turn or tip of the parabola but also offers the axis of symmetry of the graph, which is the line \(x = h\).
In our exercise, the initial function was not in vertex form, but after rewriting, we found that the vertex is at \( (0, 10) \). The axis of symmetry is the line \(x = 0\), i.e., the y-axis itself. This means our parabola is symmetrical about the y-axis and has its minimum point at \(y = 10\).Identifying the vertex helps in sketching the graph accurately and understanding the behavior of the quadratic function.
Other exercises in this chapter
Problem 66
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