Sketch the graph of f(x)=x^2/x-2. Identify the domain, x-intercepts, y intercepts, symmetry, periodicity, asymptotes, increasing and decreasing intervals, local max/min, concavity, inflexion points.
whew. can you show us what you got? Maybe you solved it correctly. Or, what isn't clear?
I'm Sort of not understanding how to find symmetry, periodicity.
The xintercept and yintercept, you plug 0 into the y to find the y intercept? then 0 into x for the xvalue? or vice versa?
Increasing and decreasing are the derivative = to 0, if its negative its decreasing and if it is positive its increasing?
Inflection points I always have a hard time on.
To sketch the graph of the function f(x) = x^2 / (x - 2), we will analyze the various characteristics of the function step by step.
1. Domain:
The function is defined for all values of x except when the denominator (x - 2) equals zero, which means x = 2. Therefore, the domain of the function is all real numbers except x = 2.
2. X-intercepts:
To find the x-intercepts, we set the function equal to zero:
x^2 / (x - 2) = 0
x^2 = 0 (since x ≠ 2)
x = 0
Thus, the x-intercept is (0, 0).
3. Y-intercept:
To find the y-intercept, we substitute x = 0 into the function:
f(0) = 0^2 / (0 - 2) = 0
Hence, the y-intercept is (0, 0).
4. Symmetry:
To determine symmetry, we evaluate whether the function is even (symmetric with respect to the y-axis) or odd (symmetric with respect to the origin). For f(x) = x^2 / (x - 2), we consider replacing x with -x:
f(-x) = (-x)^2 / (-x - 2) = x^2 / (x + 2)
Since f(-x) ≠ f(x), the function is not even. Also, since f(-x) = -f(x), it is not odd either. Thus, the function does not possess symmetry.
5. Periodicity:
To check if the function is periodic, we investigate if there exists a value 'p' such that f(x + p) = f(x) for all x. In this case, since f(x + p) = (x + p)^2 / (x + p - 2), it is apparent that the function is not periodic as changing x to x + p results in a different function.
6. Asymptotes:
Vertical Asymptotes: Vertical asymptotes occur at x = a if the function approaches infinity or negative infinity as x approaches a. For this function, there is a vertical asymptote at x = 2 since the denominator becomes zero at x = 2, resulting in division by zero.
Horizontal Asymptotes: To determine if there are any horizontal asymptotes, we investigate the behavior of the function as x goes to positive or negative infinity. For this function, as x approaches either positive or negative infinity, the function approaches y = 1, indicating that there is a horizontal asymptote at y = 1.
7. Increasing and Decreasing Intervals:
For this step, we consider the sign of the derivative of the function. The derivative, f'(x), is obtained using the quotient rule:
f'(x) = (2x(x - 2) - x^2) / (x - 2)^2
To determine the increasing and decreasing intervals, we need to find where f'(x) > 0 and f'(x) < 0. After simplifying f'(x), we find:
f'(x) = (2x^2 - 4x - x^2) / (x - 2)^2 = (x^2 - 4x) / (x - 2)^2 = x(x - 4) / (x - 2)^2
To find the critical numbers, we set f'(x) = 0:
x(x - 4) / (x - 2)^2 = 0
From this equation, we find critical points at x = 0 and x = 4.
Using these critical points, we can create a sign chart to determine the increasing and decreasing intervals:
Test x = -1: f'(-1) = (-1)(-1-4) / (-1-2)^2 = 5 / 9 (positive)
Test x = 1: f'(1) = (1)(1-4) / (1-2)^2 = -3 (negative)
Test x = 3: f'(3) = (3)(3-4) / (3-2)^2 = -3 (negative)
Test x = 5: f'(5) = (5)(5-4) / (5-2)^2 = 5 / 9 (positive)
From the sign chart, we can see that the function is increasing on the intervals (-∞, 0) and (4, ∞), and decreasing on the interval (0, 4).
8. Local Max/Min:
To find local maxima and minima, we examine the critical points found earlier:
For x = 0: f(0) = 0^2 / (0 - 2) = 0
For x = 4: f(4) = 4^2 / (4 - 2) = 8
Hence, the function has a local minimum at (4, 8), and there is no local maximum.
9. Concavity and Inflection Points:
To determine concavity and inflection points, we analyze the second derivative, f''(x).
First, let's find the second derivative:
f''(x) = (x(x - 4))'' / (x - 2)^2
After differentiating further, we get:
f''(x) = (-6x + 8) / (x - 2)^3
Setting f''(x) = 0 to find possible inflection points:
(-6x + 8) / (x - 2)^3 = 0
-6x + 8 = 0
x = 4/3
Using this value, we can perform the second derivative test:
Test x = 1: f''(1) = (-6(1) + 8) / (1 - 2)^3 = -2 (negative)
Test x = 2: f''(2) = (-6(2) + 8) / (2 - 2)^3 = undefined
Test x = 3: f''(3) = (-6(3) + 8) / (3 - 2)^3 = 2 (positive)
From the second derivative test, we see that the function changes concavity at x = 4/3. Thus, the function has an inflection point at (4/3, f(4/3)).
Now, by analyzing all the information obtained, you can sketc