Graphing Rational Functions: How to Find Asymptotes and Intercepts

[Schaus]

Homework Statement

Sketch the graphs of the following functions and show all asymptotes with a dotted line
y = (2x - 6)/ (x²-5x+4)

i) Equation of any vertical asymptote(s)

ii) State any restrictions or non-permissible value(s)

iii) Determine coordinates of any intercept(s)

iv) Describe the behavior of the function as it approaches and leaves vertical asymptotes and/or point of discontinuity

v) State the horizontal asymptote.vi) State the Domain and Range

Homework Equations

The Attempt at a Solution

I hope my thread title is correct.
First off I factored the function
y = 2(x-3)/(x-4)(x-1)

i) This gave me my vertical asymptotes: x = 4, x = 1

ii) Without any points of discontinuity then I don't have any restrictions or non-permissible values (I think)

iii) X intercept
0 = (2x-6)/(x²-5x+4)
(0)(x²-5x+4) = 2x-6
0 = 2x - 6
6 = 2x
x = 3
(3,0)

iii) Y intercept
y = 2(0)-6/(0)²-5(0)+4
y = -6/4
y = -3/2
(0,-3/2)

iv) I believe I can do this once I figure out my graph

v) Horizontal Asymptote: y = 2

vi) D: x ≠ 4, 1

Now I tried finding my range but substituting my horizontal asymptote into the function
2 = (2x-6)/(x²-5x+4)
2x²-10x+8 = 2x-6
2x²-12x+14 = 0
Using quadratic equation I get 3±√2

Now when I put all this onto my graph I don't know where to draw the lines. I think I've placed all my lines where I should but something seems wrong.

 

[ehild]

Homework Statement

Sketch the graphs of the following functions and show all asymptotes with a dotted line
y = 2x - 6/ x²-5x+4
View attachment 111366

i) Equation of any vertical asymptote(s)

ii) State any restrictions or non-permissible value(s)

iii) Determine coordinates of any intercept(s)

iv) Describe the behavior of the function as it approaches and leaves vertical asymptotes and/or point of discontinuity

v) State the horizontal asymptote.vi) State the Domain and Range

Homework Equations

The Attempt at a Solution

I hope my thread title is correct.
First off I factored the function
y = 2(x-3)/(x-4)(x-1)

i) This gave me my vertical asymptotes: x = 4, x = 1

ii) Without any points of discontinuity then I don't have any restrictions or non-permissible values (I think)

iii) X intercept
0 = 2x-6/x²-5x+4
(0)(x²-5x+4) = 2x-6
0 = 2x - 6
6 = 2x
x = 3
(3,0)

iii) Y intercept
y = 2(0)-6/(0)²-5(0)+4
y = -6/4
y = -3/2
(0,-3/2)

iv) I believe I can do this once I figure out my graph

v) Horizontal Asymptote: y = 2

vi) D: x ≠ 4, 1

Now I tried finding my range but substituting my horizontal asymptote into the function
2 = 2x-6/x²-5x+4
2x²-10x+8 = 2x-6
2x²-12x+14 = 0
Using quadratic equation I get 3±√2

Now when I put all this onto my graph I don't know where to draw the lines. I think I've placed all my lines where I should but something seems wrong.

You wrote two functions
y = 2x - 6/ x²-5x+4 which means $2x-\frac{6}{x^2}-5x+4$
and
y = 2(x-3)/(x-4)(x-1), $y=2\frac{x-3}{x-4} (x-1)$
but you seem to work with $y=\frac{2x-6}{x^2-5x+4}$
Which function is the real one?

 

[Schaus]

I see that may have been confusing. I should have written it like this
y = (2x-6)/(x²-5x+4)

 

[ehild]

I see that may have been confusing. I should have written it like this
y = (2x-6)/(x²-5x+4)

Yes, it is correct now. And it is 2(x-3)/[(x-4)(x-1)] in factorized form.
You were right, the vertical asymptotes are at x=1 and x=4. Are these x values permissible for the function? Is the function really continuous?
How is the horizontal asymptote defined? I do not think y=2 is correct.

 

[Schaus]

I was under the impression that when you had a function like 2(x-3)/[(x-4)(x-1)] the 2 in the numerator is divided by the denominator and since it is 2/1 then y = 2

 

[ehild]

I was under the impression that when xyou had a function like 2(x-3)/[(x-4)(x-1)] the 2 in the numerator is divided by the denominator and since it is 2/1 then y = 2

No, you get the horizontal asymptote at the limit x→ -∞ and x→ ∞. If x is very big the constant terms can be ignored, and the function is approximated by 2x/x²=2/x.

 

[ehild]

So, what is the horizontal asymptote?

 

[Schaus]

If its 2/x would you make 2/x = 0?

 

[ehild]

If its 2/x would you make 2/x = 0?

yes, y=0 is the horizontal asymptote.

 

[Ray Vickson]

If its 2/x would you make 2/x = 0?

No. You have the correct asypmptote, but have expressed it incorrectly.

You NEVER can have 2/x = 0; however, you do have that 2/x approaches 0 for large |x|, so $\lim_{x \to \pm \infty} 2/x = 0$ would be the right way of saying it.

 

[Schaus]

So if y = 0 then my range is y ≥ 0 because it doesn't cross the horizontal asymptote. I still don't know how to draw my graph. How do I find out which way to go with it?

 

[ehild]

So if y = 0 then my range is y ≥ 0 because it doesn't cross the horizontal asymptote. I still don't know how to draw my graph. How do I find out which way to go with it?

No. Determine the sign of the function when x<1. Remember you obtained y=-3/2 as y intercept.
Determine the sign of the function in the various domains.

 

[Schaus]

Ok, so now I start my graph from underneath the horizontal asymptote and go through the y intercept then curve it downwards -∞?

 

[Schaus]

Also when it says to state non-permissibles. Does that only mean when there are points of discontinuity? Or am I supposed to state the restrictions on x as well? Like x ≠ 4, 1

 

[ehild]

Ok, so now I start my graph from underneath the horizontal asymptote and go through the y intercept then curve it downwards -∞?

Yes. And how will the function look at the other side of the x=1 asymptote?

 

[Schaus]

If it goes down to -∞ then it will start at the top of the graph coming from ∞ then going through my (3,0) intercept down through -∞ and back to ∞ on the right side of x=4 asymptote and curving towards the right but never crossing y = 0?

 

[Schaus]

Is this how it should look? Because I was looking through my notes and it says that if you have an even exponent then the graph will look different?

 

[haruspex]

View attachment 111391 Is this how it should look? Because I was looking through my notes and it says that if you have an even exponent then the graph will look different?

Looks good to me.
Here's a handy trick: p(x)/q(x) = p(x)q(x)/q(x)², so, where defined, has the same sign as p(x)q(x). Listing all the roots of pq in order quickly shows the domains in which it is positive and which negative.

 

[ehild]

View attachment 111391 Is this how it should look? Because I was looking through my notes and it says that if you have an even exponent then the graph will look different?

It is correct. Remember, at great x values, both positive and negative, the function looks similar to 2/x (even exponent in the denominator but you have x in the numerator, too.)
Now: what is the range of the function?

 

[Schaus]

Range has to be y ∈ R because it crosses the horizontal asymptote, I believe.

 

[ehild]

Range has to be y ∈ R because it crosses the horizontal asymptote, I believe.

Yes y takes all real values. It has y=0 as asymptote, but takes the value y=0 at x=3.

 

[Schaus]

Thanks for all your help!