# Math Help - Maths function

1. ## Maths function

Consider the function f defined by f(x) =
7(x2 − 9) / ((x + 5)(x − 3)|x − 1|)
(a) What are the vertical asymptotes?
(b) What are the horizontal asymptotes?
(c) Where is the function not continuous?
(d) What are the x-intercepts and what is the y-intercept?

(a) x= -5, 3 and 1
(b) y=14,y=-28
(c) x=-5,x=3,x=1
(d) x-intercepts: x=-5,x=3,x=1.
y-intercepts: y=14,y=-28

Are these corrent??? thanks

2. Originally Posted by quah13579
Consider the function f defined by f(x) =
7(x2 − 9) / ((x + 5)(x − 3)|x − 1|)
(a) What are the vertical asymptotes?
(b) What are the horizontal asymptotes?
(c) Where is the function not continuous?
(d) What are the x-intercepts and what is the y-intercept?

(a) x= -5, 3 and 1
(b) y=14,y=-28
(c) x=-5,x=3,x=1
(d) x-intercepts: x=-5,x=3,x=1.
y-intercepts: y=14,y=-28

Are these corrent??? thanks

You are incorrect on several of these items and it is impossible for us to help you if you do not supply any work to know where you went wrong.

a) There is not a vertical asymptote at 3 because it is in fact a removable discontinuity recall $x^2-9=(x+3)(x-3)$. So the only vertical asymptotes are at -5 and 1.

b) To find the horizontal asymptotes you need to evaluate the limit of this function as $x\rightarrow \pm \infty$, you should multiply these polynomials out and notice the degree in the denominator is 3 while in the numerator it is 2, this will tell you this function approaches 0 for large values of x. You can apply L'hospital or multiply both top and bottom by $\frac{1}{x^3}$ and then take the limit to see why it is 0.
You should excercise some care if you need to be rigorous with the |x-1| when taking these limits for large positive x, |x-1|=x-1; however for large negative numbers |x-1|=-(x-1)=-x+1.

c)this is correct, I will give you the benefit of the doubt and assume it is for the right reasons.

d) These are incorrect. To find the y intercept, you evaluate at x=0. I got $\frac{7(-9)}{(5)(-3)(1)}=\frac{-63}{-15}=\frac{21}{5}=4.2$.

Now to find the x intercept you want to know for what x values is f(x)=0. This only happens when the numerator is 0. The possible points are $\pm 3$. If you recall, 3 was a removable discontinuity, so it can not be a zero, in fact even if you filled it in with a continuous function here it would be $\frac{21}{8}$. The other point though at -3 is really a zero of this function, so this is an x intercept, in fact this is the only one.