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Chapter 2: Problem 86

Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function. $$y=-4 x^{2}+20 x+160$$

Short Answer

Expert verified
The vertex of the parabola \( y=-4 x^{2}+20 x+160 \) is at (2.5, 152.5)

Step by step solution

01

Determine the vertex

In order to find the vertex (h, k), we use the formula \( h = -\frac{b}{2a} \) where a and b are the coefficients of \( x^{2} \) and x, respectively, in the given quadratic equation. For the function \( y=-4 x^{2}+20 x+160 \), a=-4 and b=20. Substituting these values into the formula, we get \( h = -\frac{20}{2(-4)} = 2.5 \). To find k (the y-coordinate of the vertex), we substitute h into the original function: \( k = -4(2.5)^{2}+20(2.5)+160 = 152.5 \). Therefore, the vertex of the parabola is (2.5, 152.5).
02

Determine a reasonable viewing rectangle and draw the graph

The vertex will be the peak of the parabola and also the point of symmetry. Knowing this, we aim to get a viewing rectangle which ensures our graph does cover at least from x=0 to x=5, and from y=150 to y=160. Using a graphing utility, the quadratic equation \( y=-4x^{2}+20 x+160 \) should be entered, and then set the window as [0, 5] x [150, 160].
03

Sketch the graph

After setting the appropriate window in your graphing utility, a downward-opening parabola should be visible with the vertex at (2.5, 152.5). This verifies our solution in Step 1.

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