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Chapter 3: Problem 84

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 given parabola \(y=5x^{2}+40x+600\) is at (-4, 500). A reasonable viewing rectangle on a graphing utility would be \([-10, 2]\) on the x-axis and \([0, 600]\) on the y-axis. The function can then be graphed using this viewing rectangle to show a downward opening parabola.

Step by step solution

01

Vertex Calculation

A vertex of a parabola in the form \(y=ax^2+bx+c\) can be found using the formula \((-b/(2a),f(-b/2a))\). The x-coordinate of the vertex is found by the formula \(-b/(2a)\). So for the given function, \(a=5\), \(b=40\), hence the x-coordinate of the vertex is \(-40/(2*5)=-4\). Substitute \(-4\) for \(x\) in the equation to get the y-coordinate of the vertex. Hence the vertex is \(-4,5*(-4)^2+40*(-4)+600=( -4,500)\).
02

Choosing the Viewing Rectangle

Once we have the vertex, we need to choose a viewing rectangle that would nicely capture the graph of the function. A good starting point for the rectangle might be a bit broader than the x-coordinate of the vertex and range of y-values from 0 to the y-coordinate of the vertex plus some additional space. A reasonable viewing rectangle might be \([-10, 2]\) on the x-axis and \([0, 600]\) on the y-axis.
03

Graphing the Function

Now we graph the function using the viewing rectangle determined in the previous step. The graph will show a downward opening parabola with the vertex at \((-4,500)\). Without a graphing utility, it would be impossible to draw the graph in this solution. However, this can be easily done using a graphing tool by inputting the function and setting the viewing rectangle as identified.

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