Asked by Ashley
what is the rational equation of a function with intercepts at (8,0) and (0,0.16) and a horizontal asymptote at y=0.5 and a vertical asymptote at x=3 and a removable discontinuity at (3,-1)
Answers
Answered by
Damon
hits y axis at x = 8
hits x axis at Y = .16
y = m x + b
when x = 0, y = .16
so
y = m x + .16
now when y = 0 , x = 8
so
0 = 8 m + .16
so m = - .16/8 = -.02
so
y = -.02 x + .16
or
100 y = -2 x + 16
hits x axis at Y = .16
y = m x + b
when x = 0, y = .16
so
y = m x + .16
now when y = 0 , x = 8
so
0 = 8 m + .16
so m = - .16/8 = -.02
so
y = -.02 x + .16
or
100 y = -2 x + 16
Answered by
Steve
You cannot have a vertical asymptote at x=3 and a removable discontinuity at x=3.
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