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Chapter 3: Problem 85

Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function. $$y=0.01 x^{2}+0.6 x+100$$

Short Answer

Expert verified
The vertex of the quadratic function \(y=0.01x^{2}+0.6x+100\) is \(-30, 89\). A reasonable viewing rectangle in the graph would be from \(-50\) to \(-10\) on the x-axis and from \(80\) to \(110\) on the y-axis.

Step by step solution

01

Identify the coefficients

This is a quadratic function in the form \(y=ax^{2}+bx+c\). The coefficients here are \(a=0.01\), \(b=0.6\) and \(c=100\).
02

Find the Vertex

The x-coordinate of the vertex can be found using the formula \(-b/2a\), which gives \((-0.6)/(2*0.01) = -30\). Then substitute \(x=-30\) into the equation to find the y-coordinate. This gives \(y=0.01*(-30)^{2}+0.6*(-30)+100 = 89\). So the vertex is \(-30, 89\).
03

Find the Viewing Rectangle

The viewing rectangle should sufficiently include the vertex and paste an approximation of the curve. A reasonable choice may be determining the x values will span from \(-50\) to \(-10\) and the y values will span from \(80\) to \(110\).
04

Graph the quadratic function

Using the graphing utility and the determined viewing rectangle, the quadratic function can be graphed plotting the parabola.
05

Interpret the Graph

From the Graph, it can be verified that the vertex of the parabola is indeed \(-30, 89\), and the parabola opens upwards as \(a\) which is \(0.01>0\).

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