Question 1166096
ax^2+bx+c=f(x)
ordered pair 1 is (0, 2)
so c=2
for 2 it is (60, 20)
so 3600x+60b+2=20
the value of x=-b/2a at the vertex so that is 60=-b/2a
-120a=b
therefore -30b+60b+2=20
30b=18
b=3/5
a=-1/200
and (-1/200)x^2+(3/5)x+2=f(distance) ANSWER
(0, 2) works
(60, 20) works
and vertex x value is -b/2a or -3/5/-1/100 or 60 for x, which works

{{{graph(300,300,-10,130,-100,30,(-1/200)x^2+(3/5)x+2)}}}