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What is the equation of the original function in the question attached here? Could you...
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To determine the function `y=1/(f(x))` depicted on the graph we have the following data:
Vertical asymptote: x=-3
Because there is an asymptote at x=-3 the denominator is zero when x=-3, thus
Substitute -3 in f(x)
Verify `y=1/(x+3)` for point (-2,1)
Therefore the given function is `y=1/(f(x))=1/(x+3)` (blue)
The original function `f(x)` is `y=x+3` (black)
The vertical asymptotes are x=-1 and x=-5, therefore the function may be:
Verify for point (-3,1/2)
Therefore the function needs to be multiplied by -2:
Verify for (-3,1/2)
Thus the given function is `y=(-2)/(x^2+6x+5)` or `y=1/(-0.5x^2-3x-2.5)` blue
The original function f(x) is `y=-0.5x^2-3x-2.5`
To write it in the form `y=a(x-h)^2+k`, first factor -0.5
Add and subtract 4 to the second factor to complete a square.
Group the terms forming a square trinomial.
Therefore the original function is `y=-0.5(x+3)^3+2` ( black)
Posted by flbyrne on October 31, 2013 at 9:28 PM (Answer #1)
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