What is the equation of the original function in the question attached here? Could you please sketch its graph?

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flbyrne | (Level 3) Assistant Educator

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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

Point: (-2,1)

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)






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