You have the following function for the height of debris:
[tex]y=-0.04x^2+2x+8[/tex]In order to answer the given questions, consider that the graph of the given function is:
1) The distance at which the debris land is given by one of the zeros of the function, because y is a parabolla, as you can notice in the previous image.
Use the quadratic formula to find the zeros of the function:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]coefficients a, b and c are the coefficients of the quadratic function ax^2+bx+c. In this case, by comparing with the given expression for y, you have:
a = -0.04
b = 2
c = 8
Replace the previous values into the quadratic formula and simplify:
[tex]\begin{gathered} x=\frac{-2\pm\sqrt[]{(2)^2-4(-0.04)(8)}}{2(-0.04)} \\ x=\frac{-2\pm\sqrt[]{4+1.28}}{-0.08} \\ x=\frac{-2\pm\sqrt[]{5.28}}{-0.08} \\ x=\frac{-2\pm2.297825059}{-0.08} \end{gathered}[/tex]select only the positive result because it has physical meaning:
[tex]x=\frac{-2-2.297825059}{-0.08}=53.72281323[/tex]Hence, the debris are at a distance of 53.72281323 feet from the launching site.
2) The maximum height reached by fireworks, is represented by the y-coordinate of the vertex point of the parabolla (as you can notice in the graph).
The x-coordinate of the vertex of a parabolla is given by:
[tex]x=-\frac{b}{2a}[/tex]Replace the values of a and b, and simplify:
[tex]x-=-\frac{2}{2(-0.04)}=25[/tex]The value of y for the previous value of x is:
[tex]\begin{gathered} y=-0.04(25)^2+2(25)+8 \\ y=-0.04(625)+50+8 \\ y=-25+50+8 \\ y=33 \end{gathered}[/tex]The previous value is the y-coordinate of the vertex.
Hence, the maximum height reached by the fireworks is 33 feet
