Chapter 10 - Section 10.1 - The Parabola and the Circle - Exercise Set - Page 612: 98

$y=-.016x^2+1.6x$

$y=ax^2+bx+c$ At 50 meters, the parabola is at its highest point of 40 meters At 0 meters, the parabola is at its lowest point of 0 meters $x=50$, $y=40$ $y=ax^2+bx+c$ $40=a*50^2+b*50+c$ $40=2500a+50b+c$ $x=0$, $y=0$ $y=ax^2+bx+c$ $0=a*0^2+b*0+c$ $0=0*a+0*b+c$ $0=c$ $40=2500a+50b+c$ $40=2500a+50b+0$ $40=2500a+50b$ $x=100$, $y=0$ $y=ax^2+bx+c$ $0=a*100^2+100*b+c$ $0=10000a+100b+c$ $0=10000a+100b$ $40=2500a+50b$ $0=10000a+100b$ $40*4=4*(2500a+50b)$ $160=10000a+200b$ $0=10000a+100b$ $160=10000a+200b$ $0-160=10000a+100b-10000a-200b$ $-160=-100b$ $-160/-100 =-100b/-100$ $16/10= b$ $8/5 =b$ $40=2500a+50b+c$ $40=2500a+50b$ $40=2500a+50*8/5$ $40=2500a+80$ $40-80=2500a+80-80$ $-40 =2500a$ $-40/2500 =2500a/2500$ $-4/250 =a$ $-4*4/250*4 =a$ $-16/1000=a$ $-.016=a$ $y=ax^2+bx+c$ $c=0$, $b=1.6$, $a=-.016$ $y=-.016x^2+1.6x$