How do you find the domain, identify any horizontal, vertical, and slant (if possible) asymptotes and identify holes, x-intercepts, and y-intercepts for #(2x^2-5x+6)/(x+2)#?

1 Answer
Jun 13, 2015


This is a complicated answer.

Asymptotes: #x = -2#, and #y = 2x - 9#
Domain: All real numbers except #-2#
Holes: None
Intercepts: y = 3 (no x-intercept)

Please read the whole explanation if you have time.


Let's start with asymptotes. To find the horizontal and vertical asymptotes of a graph, we need to see if there are points where we can divide something by 0. We can't do that with the 'y' side, but we can with the 'x' side. If 'x' is -2, then the denominator of this function will be 0, so #x = -2# is an asymptote.

Are there any more asymptotes? Well, yes, there's one more. How do I know? The degree of the numerator of this function is one more than the degree of the denominator. To find the slant asymptote of this line, we need to divide this "fraction".

#(2x^2 - 5x + 6) = (2x^2 + 4x) + (-9x - 18) + 24#

#(2x^2 - 5x + 6) = (x+2)(2x-9) + 24#

#(2x^2 - 5x + 6)/(x+2) = 2x - 9 + 24/(x+2)#

As 'x' approaches infinity, #24/(x+2)# gets closer and closer to 0, which means this function approaches the line #y = 2x - 9#.

Next, we have to find the domain. Since this function has a slant asymptote, the domain must be all real numbers except the number along the vertical asymptote, which is #-2#.

Now, we can find any holes in the equation. These are where both the numerator and denominator of the equation equal 0, so the equation has no real value at that point. The denominator will only be 0 if x is -2, so our only chance of having a hole in the function is if the numerator is also 0 if x is -2.

#2(-2)^2 - 5(-2) + 6#
#8 + 10 + 6 = 24#

Nope. This equation is free of holes.

Finally, we need to find the intercepts of the equation. To find them, we need to plug in 0 for x and y. This will give us the intercept of the opposite co-ordinate.

#y = (2(0)^2 - 5(0) + 6)/(0+2)#

#y = 6/2 = 3#

So, the y-intercept is 3.

#0 = (2x^2 - 5x + 6)/(x+2)#

#0 = 2x^2 - 5x + 6#

I can use the quadratic formula on this, or I can just tell you right now there's no real roots. This function doesn't have an x-intercept.

So, there you go! That's how to find the asymptotes, domain, holes, and intercepts of this function. I'll finish off by showing a graph of this equation so you can see its properties on a coordinate plane.
graph{y = (2x^2 - 5x + 6)/(x+2) [-32, 32, -70.7, 66.7]}