Find polynomials p and q such that p/q is undefined at 1 and 2, but p/q has a vertical asymptote only at 2. Sketch the graph of the function

Question

Please, i got an exam tmr and i have to solve this asap

Answer 1

p/q represents a rational expression.  Since this expression is undefined at x=1 and x=2, this is where the denominator cannot be zero.

 

q = (x - 1)(x - 2)

 

 

The vertical asymptote only occurs at x=2.  So x=2 is actually a discontinuity.  Therefore,

 

p = x - 1

 

 

Put this all together, we get

 

p / q = (x - 1) / (x - 1)(x - 2)

 

 

 

To graph, we use the following information:

 

Hole at x=1  (draw an open circle at this point)

Vertical asymptote is the x-axis

There is no x-intercept

y-intercept is the value of the expression at x=0

 

 

To find the endpoint behaviors, find these limits:

 

lim   p/q

x->2^-

 

In the case it increases, the curve is above the x-axis approaching from x=negative infinity.

 

In the case it decreases, the curve is below the x-axis approaching from x=negative infinity.

 

 

 

 

lim   p/q

x->2⁺

 

In the case it increases, the curve is below the x-axis approaching approaching x=positive infinity.

 

In the case it decreases, the curve is above the x-axis approaching x=positive infinity.

 

 

 

If you have any questions, let me know.

 

Answer 2

The vertical asymptote is the x-value when the denominator = 0

The vertical asymtopes are also undefined.

An instance where one would not be an asymptote is 0/0 or they cancel out when you simplify the expression. So polynomial p could be anything like: (x - 1)(x - 5) as long as it has a zero at x = 1

q = (x - 1)(x - 2)

 

p/q = (x-1)(x-5) / (x-1)(x-2)