For the function y=4-2x^2+1/6x^4 find the following:

Domain
x and y intercepts
Vertical asymptotes
Horizontal asymptotes
Symmetry
F'(x)
Critical numbers
Increasing f(x)
Decreasing f(x)
Extrema
F"(x)
Possible points of inflection
Concave up
Concave down
Points of inflection

I got some:
domain: (-infinity, infinity)
y-intercept: y=4
Symmetry: y-axis
F'(x): 2/3x^3-4x
Critical numbers: x=0, x=sqrt6, x=-sqrt6
F"(x): 2x^2-4
Possible points of inflection: (sqrt2, 2/3) (-sqrt2, 2/3)
Concave up: (-infinity, -sqrt2) and (sqrt2, infinity)
Concave down: (-sqrt2, sqrt2)

I figured out some more I think

Vertical asymptotes: none
Horizontal asymptotes: none
Extrema: relative maximum f(0)=4, relative minimums f(-sqrt2)=-2 and f(sqrt2)=-2

Would I need absolute extrema? There aren't any end points.

To find the domain of the function y = 4 - 2x^2 + (1/6)x^4, we need to determine the values of x for which the function is defined. Since this is a polynomial function, the domain is all real numbers, (-∞, ∞).

To find the x-intercepts, we set y = 0 and solve for x. So, we have 0 = 4 - 2x^2 + (1/6)x^4. This equation can be solved using factoring or other methods to find the roots.

To find the y-intercept, we set x = 0 and find the corresponding value of y. In this case, when x = 0, y = 4.

To find the vertical asymptotes, we look for any values of x that make the denominator of the function equal to zero. Since this is a polynomial function, there are no vertical asymptotes.

To find the horizontal asymptotes, we examine the behavior of the function as x approaches positive and negative infinity. The degree of the highest power of x in the function is 4. Therefore, as x approaches positive or negative infinity, the function also approaches infinity. Thus, there are no horizontal asymptotes.

To determine symmetry, we check if the function is the same when reflected over the y-axis (x = -x). In this case, the function is not symmetric about the y-axis, so there is no symmetry.

To find the derivative of the function, we differentiate each term using the power rule. The derivative of 4 is 0. The derivative of -2x^2 is -4x. The derivative of (1/6)x^4 is (1/6)(4)x^3 = (2/3)x^3. Combining these results, we have f'(x) = 0 - 4x + (2/3)x^3 = (2/3)x^3 - 4x.

To find the critical numbers, we set f'(x) = 0 and solve for x. In this case, we solve the equation (2/3)x^3 - 4x = 0. You have correctly found the critical numbers as x = 0, x = √6, and x = -√6.

To determine where the function is increasing, we look at the intervals between the critical numbers. We can use the first derivative test or the sign chart method to analyze the behavior of the derivative. For example, if we choose a value less than -√6, such as -2, we find that f'(-2) is negative, indicating that the function is decreasing. Similarly, if we choose a value between -√6 and 0, such as -1, we find that f'(-1) is positive, indicating that the function is increasing. Continuing with this analysis, we can determine the intervals where the function is increasing.

To determine where the function is decreasing, we use the same process as above, but analyze the intervals where the derivative is negative.

To find the extrema (maximums or minimums) of the function, we examine the behavior of the derivative. If the derivative changes sign from positive to negative or vice versa, it indicates a possible extrema. By analyzing the intervals using the first derivative test or the sign chart method, we can locate the values of x where the extrema occur.

To find the second derivative of the function, we differentiate the derivative function f'(x) = (2/3)x^3 - 4x. The second derivative is found by differentiating each term of the derivative function with respect to x. The derivative of (2/3)x^3 is 2x^2, and the derivative of -4x is -4. Thus, f''(x) = 2x^2 - 4.

To determine the possible points of inflection, we need to find the values of x where the second derivative changes sign. By analyzing the intervals using the second derivative test or the sign chart method, we can identify the intervals where the concavity changes and locate the possible points of inflection.

To determine where the function is concave up, we look for the intervals where the second derivative is positive.

To determine where the function is concave down, we look for the intervals where the second derivative is negative.

To find the points of inflection, we examine the behavior of the second derivative. If the second derivative changes sign from positive to negative or vice versa, it indicates a possible point of inflection. By analyzing the intervals using the second derivative test or the sign chart method, we can locate the values of x where the points of inflection occur.

You have correctly determined some of the information for the given function. Make sure to carefully analyze the intervals and signs to accurately identify the properties of the function.