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Chapter 3: Problem 82

Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function. $$y=-0.25 x^{2}+40 x$$

Short Answer

Expert verified
The vertex of the parabola \(y=-0.25x^2+40x\) is at (80, 1600). A reasonable viewing rectangle for the graphing utility is from x=0 to x=160 and y=0 to y=1700.

Step by step solution

01

Find the vertex

To find the vertex of the parabola \(y=-0.25x^2+40x\), use the formula \(-b/(2a)\) for the x-coordinate of the vertex. Here, \(a\) is -0.25 and \(b\) is 40. The formula gives the x-coordinate of the vertex as \(x=(-40)/(2*-0.25)=80\). To find the y-coordinate of the vertex, substitute \(x=80\) into the equation of the parabola to get \(y=-0.25*(80)^2+40*80=40*80-(80)^2/4\). This simplifies to \(y=1600\). So the vertex is at (80, 1600).
02

Define a viewing rectangle

A reasonable viewing rectangle can be defined as follows: because the x-coordinate of the vertex is 80, make the x-range from 0 to 160. Because the y-coordinate of the vertex is 1600 and the parabola is facing downwards, make the y-range from 0 to 1700.
03

Graph the function

Enter the function \(y=-0.25x^2+40x\) into the graphing utility. Set the viewing rectangle as described in the previous step and graph the function. The parabola should have its vertex at (80, 1600) and be facing downwards. The view rectangle should show the entire curve of the parabola from x=0 to x=160 and y=0 to y=1700.

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