Problem 82
Question
Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function. $$ y=-0.25 x^{2}+40 x $$
Step-by-Step Solution
Verified Answer
The vertex of the quadratic equation \(y=-0.25x^2 + 40x\) is at (80,400).
1Step 1: Identify the Standard Form and Formula for Vertex
The standard form of a quadratic equation is \(y = a(x-h)^2 + k\), where (h,k) is the vertex. In this exercise, the given equation is already in this form: \(y = -0.25x^2 + 40x = -0.25(x^2 - 160x)\). The vertex for a parabola represented as \(y = ax^2 + bx + c\) is given by the formula \((-\frac{b}{2a}, f(-\frac{b}{2a}))\).
2Step 2: Calculate the Vertex
To find the coordinates of the vertex, let's plug our coefficients a and b (-0.25 and 40, respectively) into the vertex formula. This gives us x-coordinate of the vertex as \(\frac{-b}{2a} = -\frac{40}{2*(-0.25)}= 80\). Then we substitute x=80 into the given equation to find the y-coordinate: \(y=-0.25 * 80^2 + 40*80 = 400\).
3Step 3: Graphing the Function
With the vertex at (80,400), one can now proceed to graph the function using a graphing utility. Ensure to use a reasonable viewing rectangle to fully capture the shape of the graph. The range and domain could be set from -100 to 500 considering the y-intercept at 0 and the vertex at 400.
Key Concepts
Vertex of a ParabolaStandard Form of a Quadratic EquationQuadratic Function Graphing Utility
Vertex of a Parabola
Understanding the vertex of a parabola is crucial when graphing quadratic functions as it represents the highest or lowest point on the graph, depending on the parabola's orientation. In a quadratic function of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, the vertex \((h, k)\) can be determined by utilizing the vertex formula \(h = -\frac{b}{2a}\) and \(k = f(h)\). In our exercise example, the quadratic function is \(y = -0.25x^2 + 40x\).
By applying the vertex formula, we can find the x-coordinate of the vertex: \(h = -\frac{40}{2 \times (-0.25)} = 80\). Substituting \(x = 80\) back into the original equation gives us the y-coordinate of the vertex: \(y = -0.25 \times 80^2 + 40 \times 80 = 400\). Therefore, the vertex of this parabola is \((80, 400)\), which will be the peak of the parabola as the coefficient \(a\) is negative, indicating the parabola opens downwards.
By applying the vertex formula, we can find the x-coordinate of the vertex: \(h = -\frac{40}{2 \times (-0.25)} = 80\). Substituting \(x = 80\) back into the original equation gives us the y-coordinate of the vertex: \(y = -0.25 \times 80^2 + 40 \times 80 = 400\). Therefore, the vertex of this parabola is \((80, 400)\), which will be the peak of the parabola as the coefficient \(a\) is negative, indicating the parabola opens downwards.
Standard Form of a Quadratic Equation
A quadratic equation is often expressed in the standard form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are coefficients, and \(x\) and \(y\) are variables. This standard form is invaluable as it makes it straightforward to identify the direction of the parabola (concave up or down), the y-intercept \(c\), and to apply the vertex formula. In the exercise, the function is nearly in standard form, but can be rewritten as \(y = -0.25(x^2 - 160x)\), which is equivalent to the original equation. Understanding this form is essential to find the vertex, identify the axis of symmetry, and graph the entire function. Once the equation is in this standard form, we can analyze key features of the parabola, such as the opening direction determined by the sign of \(a\) and the y-intercept. In this case, since \(a\) is negative, the parabola opens downwards, and since there is no constant \(c\), the y-intercept is \((0, 0)\).
Quadratic Function Graphing Utility
To graph a quadratic function, utilizing a graphing utility can greatly simplify the process and provide a visual representation of the function. A graphing utility enables students to insert the quadratic equation and automatically plots the graph. For the exercise function \(y = -0.25x^2 + 40x\), we use the vertex \((80, 400)\) to guide us in setting up a reasonable viewing rectangle on the graphing utility. This means picking a range for the x-values and y-values that comfortably include the vertex and give a full view of the parabola's shape. A suggested range might be from 0 to 160 for x (twice the x-coordinate of the vertex) and from 0 to 450 for y (slightly above the y-coordinate of the vertex). By using these guidelines, your graph will display the key features of the quadratic function: the vertex, axis of symmetry, and intercepts, making it easier to understand the function's properties and behavior.
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