# State the vertical asymptote(s), x - intercept(s), y-intercept(s), domain and range of:   f(x) =  1  4x2 - 1

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should notice that the function is not defined for any of the roots of denominator, hence, you need to find these roots, if any, such that:

`4x^2 - 1 = 0 =gt 4x^2 = 1 =gt x^2 = 1/4 =gt x_(1,2) = +-sqrt(1/4)`

`x_(1,2) = +- 1/2`

You should evaluate the following limit such that:

`lim_(x-gt1/2) 1/(4x^2 - 1) = 1/(4*(1/4) - 1) = 1/0 = oo`

Since `lim_(x-gt1/2) 1/(4x^2 - 1) `  yields `oo` , hence, the line `x = 1/2`  represents one vertical asymptote.

Evaluating the limit `lim_(x-gt-1/2) 1/(4x^2 - 1)`  yields `oo` , hence, the line `x = -1/2`  represents other vertical asymptote.

Putting f(x) = 0, you may evaluate x intercepts such that:

`1/(4x^2 - 1) = 0 `

Since denominator `4x^2 - 1!=0`  and `1!=0` , then, the function does not intercept x axis.

Putting x = 0, you may evaluate y intercepts such that:

`f(0) = 1/(4*0^2 - 1) = 1/(-1) =gt f(0) = -1`

Hence, evaluating the y intercept of the function yields (0,-1).

The domain of the function represents the assembly of x values for the function exists.

Since the denominator of function cancels for `x=+-1/2` , hence, the domain needs to exclude these values.

The range of function is the real set R.

The picture below represents the graph of the given function where you may see the vertical asymptotes and y intercept.

Sources:

lkballer24 | Student, Grade 11 | (Level 1) Valedictorian

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Sorry, these are the right figures: f(x)= 1 / 4x²-1

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