# Curve Sketching for f(x)=6x^2/(x^2-2x-15)

• Cuisine123
In summary: You can test values between -3 and 5, as well as values outside of that interval. Points of inflection occur where the concavity changes, so you can look for where y'' = 0 or is undefined (but be careful not to include x = -3 or x = 5 in this case). In summary, we discussed the components of the rational function f(x)=6x^2/(x^2-2x-15), including the intervals of increase/decrease, local maximum and minimum values, intervals of concave up/concave down, and points of inflection. We also discussed the critical numbers and how a local minimum point can have a larger y value than a local maximum point. When graph
Cuisine123

## Homework Statement

Please state the following components of this rational function?
f(x)=6x^2/(x^2-2x-15)

-intervals of increase/decrease
-local maximum and minimum values
-intervals of concave up/concave down
-points of inflection

N/A

## The Attempt at a Solution

There is a horizontal asymptote at y=6.
There are vertical asymptotes at x=-3 and x=5
The first derivative is -12x(x+15)/(x^2-2x-15)^2
The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3
According to the first derivative, the critical numbers would be -15, and 0.
After the first derivative test, I found out that a min. point occurs when x=-15 and a max. point occurs when x=0. However, I got (-15,5.625) and (0,0). How can the min. point have a larger y-value than the max. point?
Also, when I graphed this function on GraphCalc, it seems that there does not seem to be a minimum point at x=-15. Why is that?
When I graphed the function, there seems to be 3 branches; in the left, middle, and right. The left branch seems to go through the horizontal asymptote (y=6) but then it slowly approaches it from below y=6.
Please graph f(x)=6x^2/(x^2-2x-15) at http://math.ucalgary.ca/undergraduate/webwork/graphing-calculator to see what I am describing. I cannot justify why there would be a minimum at x=-15.

Last edited by a moderator:
What did you get for the derivative? The link you posted doesn't seem to have anything to do with this problem.

A local minimum point can have a larger y value than a local maximum point. Such points have smaller y values than all other points in some interval.

Mark44 said:
What did you get for the derivative? The link you posted doesn't seem to have anything to do with this problem.

A local minimum point can have a larger y value than a local maximum point. Such points have smaller y values than all other points in some interval.

Please take a look the edited information that I typed above in my first post.
I would appreciate it if you can help me out.

Cuisine123 said:

## Homework Statement

Please state the following components of this rational function?
f(x)=6x^2/(x^2-2x-15)

-intervals of increase/decrease
-local maximum and minimum values
-intervals of concave up/concave down
-points of inflection

N/A

## The Attempt at a Solution

There is a horizontal asymptote at y=6.
There are vertical asymptotes at x=-3 and x=5
The first derivative is -12x(x+15)/(x^2-2x-15)^2
Everything is fine to here. I didn't check the 2nd derivative, though.
Cuisine123 said:
The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3
According to the first derivative, the critical numbers would be -15, and 0.
After the first derivative test, I found out that a min. point occurs when x=-15 and a max. point occurs when x=0. However, I got (-15,5.625) and (0,0). How can the min. point have a larger y-value than the max. point?
These are the correct critical numbers. As noted in my other post, a local minimum point can have a larger y value than another local maximum point.
Cuisine123 said:
Also, when I graphed this function on GraphCalc, it seems that there does not seem to be a minimum point at x=-15. Why is that?
Change your scale and it will show up better. Going to the left from the y axis, the graph drops down to its local min point, and then gradually moves up the the hor. asymptote.
Cuisine123 said:
When I graphed the function, there seems to be 3 branches; in the left, middle, and right. The left branch seems to go through the horizontal asymptote (y=6) but then it slowly approaches it from below y=6.
Please graph f(x)=6x^2/(x^2-2x-15) at http://math.ucalgary.ca/undergraduate/webwork/graphing-calculator to see what I am describing. I cannot justify why there would be a minimum at x=-15.

Last edited by a moderator:
Mark44 said:
Everything is fine to here. I didn't check the 2nd derivative, though.
These are the correct critical numbers. As noted in my other post, a local minimum point can have a larger y value than another local maximum point.
Change your scale and it will show up better. Going to the left from the y axis, the graph drops down to its local min point, and then gradually moves up the the hor. asymptote.

Will I need to use the values of -15, -3, 0, and 5 for the chart to find intervals of increase and decrease?

Also, how do I find the intervals of concave up and concave down and the points of inflection?
The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3. For what values of x do I test for in the 2nd derivative chart? Are -3 and 5 two numbers that I must include in this chart, since either one of them makes the second derivative undefined? Please help.

Cuisine123 said:
Will I need to use the values of -15, -3, 0, and 5 for the chart to find intervals of increase and decrease?
Yes.
Cuisine123 said:
Also, how do I find the intervals of concave up and concave down and the points of inflection?
Concave up - where y'' > 0. Concave down - where y'' < 0.
Cuisine123 said:
The second derivative is 12(2x^3+45x^2+225)/(x^2-2x-15)^3. For what values of x do I test for in the 2nd derivative chart? Are -3 and 5 two numbers that I must include in this chart, since either one of them makes the second derivative undefined? Please help.
Your intervals of concave up/concave down can't include x = -3 or x = 5, since the function is not defined there.

## 1. What is the domain of the function f(x) = 6x^2/(x^2-2x-15)?

The domain of a function is the set of all possible input values for which the function is defined. In this case, the function is defined for all values of x except x=5 and x=-3, as these would result in a division by zero. Therefore, the domain of f(x) is all real numbers except x=5 and x=-3.

## 2. How do you find the x-intercepts of the graph of f(x) = 6x^2/(x^2-2x-15)?

The x-intercepts of a graph are the points where the graph crosses the x-axis. To find the x-intercepts of this function, we set f(x) equal to zero and solve for x. We can do this by factoring the numerator and denominator to get 6x^2/(x-5)(x+3) and then setting each factor equal to zero. This gives us x=0, x=5, and x=-3 as the x-intercepts of the graph.

## 3. What are the critical points of the function f(x) = 6x^2/(x^2-2x-15)?

The critical points of a function are the points where the slope of the graph is equal to zero. To find the critical points of this function, we first find the derivative of f(x) using the quotient rule. Then, we set the derivative equal to zero and solve for x. This gives us x=0 and x=5 as the critical points of the graph.

## 4. How can you determine the concavity of the graph of f(x) = 6x^2/(x^2-2x-15)?

The concavity of a graph indicates the direction in which the graph is curving. To determine the concavity of this function, we can find the second derivative of f(x) using the quotient rule. Then, we set the second derivative equal to zero and solve for x. This gives us x=0 as the point where the graph changes from concave up to concave down. We can also use the first derivative test to analyze the sign of the first derivative and determine the intervals where the function is concave up or concave down.

## 5. What are the asymptotes of the graph of f(x) = 6x^2/(x^2-2x-15)?

An asymptote is a line that the graph of a function approaches but never touches. To find the asymptotes of this function, we can analyze the behavior of the function as x approaches positive and negative infinity. As x approaches positive infinity, the function approaches the line y=6, and as x approaches negative infinity, the function approaches the line y=-6. Therefore, the asymptotes of the graph are y=6 and y=-6.

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