Intercept of Asymptote

Algebra Level 4

Find the y y -intercept of the oblique asymptote of the following curve.

y = x 2 + 5 x + 6 x 2 \large y = \frac{ x^{2} + 5 x + 6 } { x - 2 }


The answer is 7.

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3 solutions

Pranshu Gaba
Jan 19, 2016

Note that we can write the given equation as

y = x 2 2 x + 7 x + 6 x 2 y = \frac{ x^{2 }- 2x + 7x + 6 }{x - 2 } y = x ( x 2 ) + 7 x 14 + 20 x 2 y = \frac { x ( x - 2) + 7x -14 + 20 } { x - 2} y = x + 7 ( x 2 ) + 20 x 2 y = x + \frac {7(x - 2 ) + 20 } { x - 2} y = x + 7 + 20 x 2 y = x + 7 + \frac{ 20 } { x - 2 }

Observe that when the absolute value of x x is very very large, like for example x = 1000000 x= 1000000 or x = 3000000 x = - 3000000 , the fraction term on the right hand side becomes very small and y is approximately equal to x + 7 x + 7 . When x ± x \to\pm \infty , y x + 7 y \to x + 7 . Thus, y = x + 7 y = x + 7 is an oblique asymptote of the given curve.

The y intercept of y = x + 7 y = x + 7 is the value of y y when x = 0 x = 0 , that is y = 7 y = \boxed{ 7 } _\square

Ravi Dwivedi
Jan 23, 2016

Put y = m x + c y=mx+c in the given equation and since the equation we need is not vertical therefore we can multiply both sides by ( x 2 ) (x-2) and get ( m x + c ) ( x 2 ) = x 2 + 5 x + 6 (mx+c)(x-2)=x^2+5x+6

( m 1 ) x 2 ( 2 m c + 5 ) x ( 2 c + 6 ) = 0 \implies (m-1)x^2-(2m-c+5)x-(2c+6)=0

This equation has double root at x > x -> \infty

So put coeff of x 2 x^2 = coefficient of x x = 0 0

m 1 = 2 m c + 5 = 0 m-1=2m-c+5=0

m 1 = 0 , 2 m c + 5 = 0 \implies m-1=0,2m-c+5=0

m = 1 , c = 7 m=1,c=7

So the y-intercept is c = 7 c=\boxed{7}

Moderator note:

It's easier to use partial fractions to figure out what the graph looks like.

Kevin Yang
Nov 18, 2018

link text See this

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