The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
Yes, the tangent function is periodic.
y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!
No.
When you graph a tangent function, the asymptotes represent x values 90 and 270.
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
tan (0) = opposite/adjacent
it depends on what b is in the equation. Period = 360 degrees / absolute value of b.
Yes, the tangent function is periodic.
y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!y = e2 or e2 is not a function of x: it is a constant. So it is a horizontal straight line and its tangent, at any point, is itself.If you think I am going to sketch a graph on this browser, you have another think coming!
A tangent function is a trigonometric function that describes the ratio of the side opposite a given angle in a right triangle to the side adjacent to that angle. In other words, it describes the slope of a line tangent to a point on a unit circle. The graph of a tangent function is a periodic wave that oscillates between positive and negative values. To sketch a tangent function, we can start by plotting points on a coordinate plane. The x-axis represents the angle in radians, and the y-axis represents the value of the tangent function. The period of the function is 2Ο radians, so we can plot points every 2Ο units on the x-axis. The graph of the tangent function is asymptotic to the x-axis. It oscillates between positive and negative values, crossing the x-axis at Ο/2 and 3Ο/2 radians. The graph reaches its maximum value of 1 at Ο/4 and 7Ο/4 radians, and its minimum value of -1 at 3Ο/4 and 5Ο/4 radians. In summary, the graph of the tangent function is a wave that oscillates between positive and negative values, crossing the x-axis at Ο/2 and 3Ο/2 radians, with a period of 2Ο radians.
You need more than one tangent to find the equation of a parabola.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
A circle's tangent is exactly the same as a triangle's tangent. If you look at a circle, you can make the radius the hypotenuse. Then make a vertical line from the point, and a horizontal line from the center. If you look, you have a triangle made inside the circle. This is why angles can be measured in radians, a unit that is derived from the circumference of a circle.-------------------------------------------------------------------------------------------By doing a little calculus, we find that the slope of the equation of a circle-the slope of the tangent line-is given by the tangent of an angle.AnswerEverything written above is correct, but doesn't have anything to do with tangents (in the circle sense of the word). Suppose you're given an angle theta. Draw a circle together with two radii, one horizontal and the other at an angle theta from the first one. (So far, this is the same as above.) Now draw the tangent to the circle at X, the point where the non-horizontal radius meets the circumference. Let Y be the point where this tangent meets the horizontal line through the centre. Then, assuming the radius is 1, tan(theta) is the distance XY, which is the length of part of the tangent.
The first thing you may want to do would be to find the tangent line to the function. The tangent line is a line that passes through a given point on a function, but does not touch any other point on the function (assuming the function is one to one). Assuming you have the tangent line, the normal line is simply perpendicular to the tangent line- it forms a 90 degree angle with the tangent line. One you have the tangent line and the point which it passes through, you can find the normal line. To obtain the perpendicular line to any function, take the inverse reciprocal of the slope (if your slope was 2, it is now -.5). After that, plug in your (x, y) coordinate, and you can solve for the constant b (assuming there is one). This should give the normal line to a tangent of at a point on a function.
No.
Say you are given a function and an x value.(1) First find the y coordinate that corresponds to that x value by plugging x into the function and simplifying to find y = some #. Now you have a point (x, y) that is not only on the function, but also on the tangent line.(2) Take the derivative of the function.(3) If the derivative still has xs in it, plug in the x value you were given and simplify. This should give you an actual number--the slope of the tangent line.(4) From steps 1 and 3, you now have a point on the tangent line and the slope of the tangent line. Use these two things to write the equation for the tangent line in y=mx+b form (m is the sope, plug in the point you found, solve for b, then rewrite the equation replacing m and b but leaving in x and y).