# Thread: Sketch the function (x+5)/(x^2-16) and find.......

1. ## Sketch the function (x+5)/(x^2-16) and find.......

Hi guys i have a question out of a book that i cant seem to solve, you have to sketch the graph of (x+5)/(x^2-16) and find the following:
so far i got:

1. Domain, the domain i found to be (-inf, -4) U (-4, 4) U (4, inf)
2. Vert Asymptotes, 9/0, 1,0
3. Horz Asymptotes, no horizontal asys
4. Axial Intercepts, not sure what this means
5. Local Maxima, i dont know how to find it
6. Local Minima, "
7. Point of inflection, ?
8. Identify one interval where the function is conc up and one interval where it is conc down?

Please help there are questions in the later chapters that require you to have the knowledge of this question.

Thanks, TC

2. ## Re: Sketch the function (x+5)/(x^2-16) and find.......

Originally Posted by iFuuZe
Hi guys i have a question out of a book that i cant seem to solve, you have to sketch the graph of (x+5)/(x^2-16) and find the following:
so far i got:

1. Domain, the domain i found to be (-inf, -4) U (-4, 4) U (4, inf)
2. Vert Asymptotes, 9/0, 1,0
3. Horz Asymptotes, no horizontal asys
4. Axial Intercepts, not sure what this means
5. Local Maxima, i dont know how to find it
6. Local Minima, "
7. Point of inflection, ?
8. Identify one interval where the function is conc up and one interval where it is conc down?

Please help there are questions in the later chapters that require you to have the knowledge of this question.

Thanks, TC
The vertical asymptotes are given by the values of x that you can not take. So there are vertical asymptotes at x = -4 and x = 4.

The horizontal asymptotes are given by the values of y that you can not take. The easiest way I can think of to do this is to evaluate the inverse relation (which would make the x and y values swap) and evaluate the values that x can not take in the inverse relation.

Axial intercepts: Means what are the y intercepts and x intercepts.

Local maxima and minima (aka critical points) are found where the derivative of your function is 0. If the second derivative is positive at a critical point, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum.

Points of inflexion are where the function changes from increasing slope to decreasing slope or vice versa. In other words, points of inflexion are where the second derivative is 0.

A graph is concave up where the second derivative is positive, and concave down where the second derivative is negative.