i really need some help, i really appreciate all yalls help.
Consider the function F(x)=2x^2-5x+5/x-2
Domain
Vertical,horizontal,or slant asymptotes
x-intercept(s)
y-itercept(s)
symmetry(respect to x, y-axis or orgin)
F’ (x)
Critical numbers
F” (x)
Possible points of inflection
Can't you answer any of these? If not, you need private tutoring. That is not what we provide here.
Show us what you can do so we can see where you most need help, and someone will provide help with those areas.
Marlyn: I removed your someone post. What drwls is saying, it appears that you are answer grazing, you show now work on a variety of topics, not indicating what you don't understand about the question. Surely you know something..that is our starting point for teaching. We cant begin with "domains" and end in points of inflection in one post....that ranges from Alg II to basic calculus.
Marlyn, I do not know if you have gone back to look at my detailed answer to your post on Dec. 5.
http://www.jiskha.com/display.cgi?id=1260035879
If after reading the answer, you still have difficulties doing this current question, I must conclude that drwls is right. You need tutoring, and from pre-calculus up, not just calculus.
If you can do at least part of the current question, post what you have so we know where you need help. You cannot expect answers to multiple questions of the same type.
I'm here to help! Let's break down each question and explain how to find the answers step by step.
1. Domain:
To find the domain of a function, you need to determine all possible values of x for which the function is defined. In the given function F(x) = (2x^2 - 5x + 5)/(x - 2), the only restriction is that the denominator (x - 2) should not equal zero since division by zero is undefined. Therefore, the domain of F(x) is all real numbers except for x = 2.
2. Vertical, Horizontal, or Slant Asymptotes:
To determine the existence of asymptotes, you need to analyze the behavior of the function as x approaches positive and negative infinity. In this case, since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
To check for vertical asymptotes, set the denominator equal to zero and solve for x. In this case, x - 2 = 0, so x = 2. Therefore, there is a vertical asymptote at x = 2.
There is no slant asymptote because the degree of the numerator is not exactly one larger than the degree of the denominator.
3. x-intercepts:
To find the x-intercepts, set F(x) equal to zero and solve for x. In this case, set (2x^2 - 5x + 5)/(x - 2) = 0. However, it is important to note that the simplified form of the equation (2x^2 - 5x + 5)/(x - 2) = 0 is not possible. The reason is that the numerator, 2x^2 - 5x + 5, is irreducible and does not factor. Therefore, the given function does not have any x-intercepts.
4. y-intercepts:
To find the y-intercept, set x equal to zero and evaluate F(0). Substitute x = 0 into the function F(x) = (2x^2 - 5x + 5)/(x - 2). You will get F(0) = (2(0)^2 - 5(0) + 5)/(0 - 2) = 5/(-2) = -5/2. So the y-intercept is at the point (0, -5/2).
5. Symmetry (respect to x, y-axis, or origin):
To determine symmetry, evaluate if the function is equal when substituting -x for x (reflection about the y-axis) or y for x (reflection about the x-axis). In this case, substituting -x for x in F(x) does not produce an equivalent expression. Therefore, the function is not symmetric with respect to the y-axis.
Similarly, substituting y for x does not produce an equivalent expression. Therefore, the function is not symmetric with respect to the x-axis or origin.
6. F'(x):
To find F'(x), the first derivative of F(x), differentiate the function using the rules of differentiation. In this case, differentiate the function F(x) = (2x^2 - 5x + 5)/(x - 2) using the quotient rule or other applicable rules of differentiation. The result will be the first derivative, F'(x).
7. Critical Numbers:
Critical numbers occur when the first derivative, F'(x), is equal to zero or undefined. To find the critical numbers, solve the equation F'(x) = 0 and identify any values of x that make the first derivative undefined.
8. F''(x):
To find F''(x), the second derivative of F(x), differentiate the first derivative, F'(x), with respect to x.
9. Possible Points of Inflection:
Points of inflection occur where the concavity of the graph changes. To find the possible points of inflection, solve the equation F''(x) = 0 and identify any values of x that make the second derivative equal to zero.
Remember, these explanations are general guidelines, and the specific calculations may vary depending on the given function. It is always important to double-check your calculations and interpretations to ensure accuracy.