Use the algorithm for curve sketching to sketch the following:
y= 2x/(x^2-25)
you know that
there is an x-intercept and a y-intercept at (0,0)
there are two vertical asymptotes at x = -5 and +5
The horizontal asymptote is y=0
type in your function at wolframalpha.com to verify your sketch.
To sketch the curve of the equation y = 2x / (x^2 - 25), we can use the algorithm for curve sketching, which involves the following steps:
1. Determine the domain of the function:
The function y = 2x / (x^2 - 25) will be undefined when the denominator x^2 - 25 equals zero. Thus, we need to find the values of x for which x^2 - 25 = 0. Solving this equation, we have:
x^2 - 25 = 0
(x + 5)(x - 5) = 0
x = -5, x = 5
Therefore, the domain of the function is all real numbers except x = -5 and x = 5.
2. Determine the intercepts:
To find the y-intercept, we set x = 0:
y = 2(0) / (0^2 - 25)
y = 0
Hence, the y-intercept is (0, 0).
To find the x-intercepts, we set y = 0 and solve for x:
2x / (x^2 - 25) = 0
2x = 0
x = 0
Hence, the x-intercept is (0, 0).
3. Determine the symmetry of the function:
The function y = 2x / (x^2 - 25) is not symmetric with respect to the x-axis, y-axis, or the origin.
4. Determine the vertical asymptotes:
Vertical asymptotes occur when the denominator of a rational function becomes zero but the numerator does not. In this case, the denominator x^2 - 25 equals zero when x = -5 and x = 5 (as mentioned in step 1). Therefore, the vertical asymptotes are x = -5 and x = 5.
5. Determine the horizontal asymptotes:
To find the horizontal asymptote, we analyze the behavior of the function as x approaches positive infinity or negative infinity. For the given function, the degree of the numerator (which is 1) is less than the degree of the denominator (which is 2).
Since the degree of the denominator is greater, the horizontal asymptote is y = 0.
6. Determine the critical points:
To find the critical points, we find the values of x where the derivative of the function is equal to zero or undefined.
First, let's find the derivative of the function:
f(x) = 2x / (x^2 - 25)
f'(x) = (2 * (x^2 - 25) - 2x * (2x)) / (x^2 - 25)^2
= (2x^2 - 50 - 4x^2) / (x^2 - 25)^2
= (-2x^2 - 50) / (x^2 - 25)^2
The derivative will be undefined when the denominator (x^2 - 25)^2 equals zero. However, this never happens because the square of a real number is always positive.
Now, let's solve -2x^2 - 50 = 0 to find the x-coordinate of the critical points:
-2x^2 - 50 = 0
-2(x^2 + 25) = 0
x^2 + 25 = 0
Since x^2 is always non-negative, there are no real solutions to this equation. Therefore, there are no critical points.
7. Determine the increasing and decreasing intervals:
To find the increasing and decreasing intervals, we can use the sign of the derivative. However, since there are no critical points (where the derivative is zero), we cannot determine the intervals where the function is increasing or decreasing.
8. Sketch the curve:
With the information we have gathered, let's sketch the curve:
- Plot the x-intercept at (0, 0).
- Plot the y-intercept at (0, 0).
- Draw the vertical asymptotes at x = -5 and x = 5.
- Draw the horizontal asymptote at y = 0.
Since we cannot determine the increasing and decreasing intervals, we cannot draw the curve between the asymptotes. However, based on the behavior of the function, as x approaches the vertical asymptotes, it becomes very large in magnitude. This information can help you draw a reflection of the curve between the asymptotes.
Keep in mind that this sketch is just a rough representation and it's always a good idea to use technology or graphing software to obtain a more accurate graph.
I hope this step-by-step explanation helps you sketch the curve of y = 2x / (x^2 - 25)!
To sketch the curve of the function y = 2x / (x^2 - 25), we will follow the algorithm for curve sketching. This algorithm involves the following steps:
1. Determine the domain of the function:
The function y = 2x / (x^2 - 25) is defined for all real numbers except when the denominator, x^2 - 25, equals zero. Solving x^2 - 25 = 0, we find x = -5 and x = 5. Hence, the domain of the function is (-∞, -5) ∪ (-5, 5) ∪ (5, ∞).
2. Find the x and y intercepts:
To find the x-intercepts, we set y = 0 and solve for x. In this case, there are no x-intercepts.
To find the y-intercepts, we set x = 0 and solve for y. Substituting x = 0 into the equation, we get y = 0.
3. Determine the vertical asymptotes:
Vertical asymptotes occur when the denominator equals zero, i.e., x^2 - 25 = 0. Solving for x, we find x = -5 and x = 5. Therefore, the vertical asymptotes are x = -5 and x = 5.
4. Determine the horizontal asymptotes:
To find the horizontal asymptotes, we examine the behavior of the function as x approaches positive or negative infinity. Divide the numerator degree by the denominator degree. In this case, the numerator has degree 1 and the denominator has degree 2. Since the numerator degree is less than the denominator degree, the horizontal asymptote is y = 0.
5. Determine the symmetry:
To check for symmetry, substitute -x in place of x in the equation. Simplifying, we have y = -2x / (x^2 - 25). Since the equation is not unchanged, the function is not symmetric about the y-axis.
6. Determine the intervals of increase, decrease, and local extrema:
To find the intervals of increase and decrease, we need to find the critical points. The critical points occur when the derivative of the function is equal to zero or undefined. Taking the derivative of y with respect to x, we get:
y' = (2 * (x^2 - 25) - 2x * 2x) / (x^2 - 25)^2
= (2x^2 - 50 - 4x^2) / (x^2 - 25)^2
= (-2x^2 - 50) / (x^2 - 25)^2
Setting y' equal to zero, we have:
-2x^2 - 50 = 0
x^2 = -25 (no real solutions)
This indicates that there are no critical points, so the function is either always increasing or always decreasing within its domain.
To identify the local extrema, we can inspect the endpoints of the interval (-∞, -5), (-5, 5), and (5, ∞). However, since the function does not have any local extrema, we can skip this step.
7. Determine the concavity and inflection points:
To find the concavity, we take the second derivative of the function. The second derivative is given by:
y'' = 2 * (4x(x^2 - 25) - (-2x^2 - 50) * 2x) / (x^2 - 25)^3
= (8x^3 - 200x + 4x^3 + 100x) / (x^2 - 25)^3
= (12x^3 - 100x) / (x^2 - 25)^3
To find the points of inflection, we set y'' equal to zero and solve for x:
(12x^3 - 100x) / (x^2 - 25)^3 = 0
12x^3 - 100x = 0
4x(3x^2 - 25) = 0
x = 0 or x = sqrt(25/3) = 5√3/3 or x = -sqrt(25/3) = -5√3/3
These points divide the x-axis into four intervals: (-∞, -5√3/3), (-5√3/3, 0), (0, 5√3/3), and (5√3/3, ∞). By testing points within each interval, we can determine the concavity of the function.
8. Sketch the graph:
Based on the previous steps, we can now sketch the graph of the function. Here is the sketch:
-> Vertical asymptotes: x = -5 and x = 5
-> Horizontal asymptote: y = 0
-> No x-intercepts
-> y-intercept: (0, 0)
-> No local extrema or points of inflection
The graph should have a vertical asymptote at x = -5, a vertical asymptote at x = 5, a horizontal asymptote at y = 0, and passes through the y-intercept at the origin (0, 0). Since the function is always increasing or always decreasing, the curve will not cross the horizontal asymptote.
The graph will look similar to a hyperbola that approaches the vertical asymptotes near x = -5 and x = 5 but never crosses them.