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

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### 1 Answer

1.

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

`f(x)=x+3`

Substitute -3 in f(x)

`f(x)=-3+3=0`

Verify `y=1/(x+3)` for point (-2,1)

`1=1/(-2+3)`

Therefore the given function is `y=1/(f(x))=1/(x+3)` (blue)

**The original function `f(x)` is `y=x+3` **(black)

2.

The vertical asymptotes are x=-1 and x=-5, therefore the function may be:

`y=1/((x+5)((x+1)`

Verify for point (-3,1/2)

`y=1/((-3+5)(-3+1))=1/((2)(-2))=-1/4`

Therefore the function needs to be multiplied by -2:

`y=-2/((x+5)(x+1))`

Verify for (-3,1/2)

`1/2=-2/((-3+5)(-3+1))`

`1/2=-2/((-2)((2))`

`1/2=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

`y=-0.5(x^2+6x+5)`

Add and subtract 4 to the second factor to complete a square.

`y=-0.5(x^2+6x+9-4)`

Group the terms forming a square trinomial.

`y=-0.5(x+3)^2-0.5(-4)`

Simplify:

`y=-0.5(x+3)^2+2`

Therefore the original function is `y=-0.5(x+3)^3+2` ( black)