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Chapter 2: Problem 87

Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function. $$y=5 x^{2}+40 x+600$$

Short Answer

Expert verified
The vertex of the function \(y = 5x^2 + 40x + 600\) is at the point \(-4, 460\). A reasonable viewing rectangle for the graph would be X from \(-10\) to \(10\) and Y from \(0\) to \(500\). The graph will be an upward opening parabola.

Step by step solution

01

Find X-Coordinate of the Vertex

The x-coordinate of the vertex can be found using the formula \(-b/2a\). Here, \(a = 5\) and \(b = 40\). So, the x-coordinate of the vertex will be \(-40/ (2*5) = -4\).
02

Find Y-Coordinate of the Vertex

The y-coordinate is calculated by plugging the x-coordinate value into the function \(y = 5x^2 + 40x + 600\). So \(y = 5(-4)^2 + 40(-4) + 600 = 5*16 - 160 + 600 = 460\). Therefore, the vertex of the function is \(-4, 460\).
03

Determine a Reasonable Viewing Rectangle for the Graph

A reasonable viewing rectangle for the graph can be determined knowing that the vertex is \(-4, 460\). So, a good viewing rectangle could be x values from \(-10\) to \(10\) (encompassing \(-4\)) and y values from \(0\) to \(500\) (encompassing \(460\)).
04

Graph the Function

Enter the quadratic function \(y = 5x^2 + 40x + 600\) into a graphing utility and set the viewing rectangle as determined in Step 3. The vertex of the parabola will be at point \(-4, 460\). The parabola will be upward opening as the coefficient of \(x^2\) is positive.

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