I had trouble trying to figure out why y=x+3/(x-6)(x-1) has a x intercept of (3,0) if the horizontal asymptote is y=0. Can someone explain to me why the line crosses the x axis even if there is a vertical asymptote of y=0?
I had trouble trying to figure out why y=x+3/(x-6)(x-1) has a x intercept of (3,0) if the horizontal asymptote is y=0. Can someone explain to me why the line crosses the x axis even if there is a vertical asymptote of y=0?
If the function is $\displaystyle y= \frac{x+ 3}{(x- 6)(x- 1)}$ then (3, 0) is NOT an x- intercept. The y coordinate is, of course, 0 but then we gave $\displaystyle \frac{x+ 3}{(x- 6)(x- 1)}= 0$. Multiplying on both sides by (x- 6)(x- 1) we have $\displaystyle x+ 3= 0$ so that the x-intercept is (-3, 0).
But the fact that the horizontal asymptote is y= 0 only means that the graph approaches the x-axis as x goes to either positive or negative infinity. That says nothing about what happens for any finite value of x.
Sorry I meant to say the x intercept is (-3,0). The correct function is $\displaystyle \dfrac {x+3}{(x-6)(x-1)}$. I'm asking why the line crosses the x axis if there is a horizontal asymptote of y=0. My phone can't display the formatting so I have to type it manually.
First off, it should be y=(x+3)/(x-6)(x-1) if you are to have an x-int at (-3,0) because all x-intercepts are found in the numerator of the rational function.
Second, the graph of a line can cross a Horizontal Asymptote none, once, or many times locally and eventually flatten out along among the H.A. In other words, it is perfectly fine for this graph to have a line cross the H.A through an x-intercept at (-3,0).
Here are the steps to finding the H.A and whether line crosses H.A at any point if ever.
1) Find H.A
m<n , therefore H.A: y=0
2) Does graph cross H.A locally?
To find this we set your reduced form equation equal to the H.A value (y=0) like so and solve for x.
0 = (x+3)/(x-6)(x-1)
Multiply both sides by denominator of right side ((x-6)(x-1)) or just don't bother since anything times 0 is 0.
We get: 0 = (x+3)
Substract 3 from both sides and : x= -3
Therefore, as you can see, the graph with have a line pass through point (3,0) which happens to lie on H.A.
Please be careful with notation when trying to help on the forums. Your response is appreciated, but would be appreciated far more with more exact notation.
(x+3)/(x-6)(x-1) is ambiguous and by a strict reading does not mean what you want it to mean.
$(x+3)/(x-6)(x-1)=\dfrac{x+3}{x-6}(x-1)$
$(x-3)/[(x-6)(x-1)]=\dfrac{x+3}{(x-6)(x-1)} $
Secondly, the graph of a line only has a horizontal asymptote if the line has a slope of zero. You may have meant "the graph of a function". Please be precise with your terminology. Those requesting help do not have the lifetime of mathematical study that those who help have, so it is imperative that helpers are extremely careful to only make true statements to avoid confusing those they try to help.
Next, the step one you write where m < n may have meaning for you, but there is no indication of an m or an n in the problem. I am not sure what you mean by that, so someone asking for help is even less likely to know what you mean. Not everyone is taught math the exact same way. So, please try to keep your notation universal. I assume by m, you mean the highest power of x in the numerator and by n you mean the highest power of x in the denominator. Next time, please write it out. Again, I only suspect that is what you mean because I have a lifetime of mathematical experience to aid my comprehension.
It is not typically helpful to try to explain something the original poster did not identify as a problem. The original poster did not need a step-by-step explanation of how to find horizontal asymptotes as he/she provided the correct horizontal asymptote in the original question.