Question

Find the asymptotes of the curve (2x+3)y=(x-1)^2​

Answer 1

Given :Equation :$(2x+3)y=(x-1)^2$

To find :Find the asymptotes of the curve

Solution:

We are supposed to find the asymptotes of the curve

Equation :$(2x+3)y=(x-1)^2$

$\Rightarrow y =\frac{(x-1)^2}{(2x+3)}\\\Rightarrow y = \frac{x^2-2x+1}{2x+3}$

Vertical asymptote : Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

Equate numerator of given curve equals to 0

So,2x+3=0

$x=\frac{-3}{2}$

So, Vertical asmptote is -2/3

Horizontal asymptote :A horizontal asymptote is a y-value on a graph which a function approaches but does not actually reach.

The given polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote.

Slant asymptote:A slant (oblique) asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator.

To find the slant asymptote you must divide the numerator by the denominator

On dividing numerator by denomintaor

Quotient :$\frac{x}{2}-\frac{7}{4}$

So, Oblique asymptote :$y =\frac{x}{2}-\frac{7}{4}$

Answer 2

The vertical asymptote of the given curve is x =  $\frac{-3}{2}$ .

There is no horizontal asymptote in our function.

Step-by-step explanation:

We are given the expression of the curve:  (2x+3)y=(x-1)^2

Rearranging the given expression we get;

                   $y=\frac{(x-1)^{2} }{2x+3}$

In this the numerator is $(x-1)^{2}$ and the denominator is (2x + 3).

Now, there are mainly two types of asymptotes:

1. Vertical Asymptote
2. Horizontal Asymptote

For finding the vertical asymptote of any curve, just equate the denominator of the function with 0 and calculate for x, that means;

            2x + 3 = 0

              2x = -3

                x  =  $\frac{-3}{2}$

So, the vertical asymptote of the given curve is x =  $\frac{-3}{2}$ .

Now, for finding the horizontal asymptote there are three criteria;

• If both numerator and denominator are of same degrees polynomials, then we have to divide the coefficients of the highest degree terms from both polynomials.
• If the polynomial in the numerator is of lower degree than the denominator, then the x-axis, i.e. (y = 0) is considered to be the horizontal asymptote.
• If the polynomial in the numerator is of higher degree than the denominator, then there is no horizontal asymptote. Instead of that, there is a slant asymptote.

Now, in our case the polynomial function in the numerator is of higher degree, (i.e. 2) than the denominator, so there is no horizontal asymptote.

Hence, there is no horizontal asymptote in our function.