According to Wikipedia.
'Suppose that a curve is given as the graph of a function, y = f(x). To find the tangent line at the point p = (a, f(a)), consider another nearby point q = (a + h, f(a + h)) on the curve. The slope of the secant passing through p and q is equal to the difference quotient
As the point q approaches p, which corresponds to making h smaller and smaller, the difference quotient should approach a certain limiting value k, which is the slope of the tangent line at the point p. If k is known, the equation of the tangent line can be found in the point-slope form:'
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.
That line is [ y = 2 cos(2x) ].
a tangent is a line that touches the circle at only ONE point
For the equation (9x^2)/(x^2+4)
To measure the point at which two tangents intersect each other, find an equation for each tangent line and compute the intersection. The tangent is the slope of a curve at a point. Knowing that slope and the coordinates of that point, you can determine the equation of the tangent line using one of the forms of a line such as point-slope, point-point, point-intercept, etc. Do the same for the other tangent. Solve the two equations as a system of two equations in two unknowns and you will have the point of intersection.
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.
The question is suppose to read: Find the equation of the line tangent to y=(x²+3x)²(2x-2)³, when x=8
find the derivative to get the slope, then using the graph of the other line given to you, find a point on the graph that they share and plug it in to find the y-value. then use the point slope formulahttp://docs.google.com/gview?a=v&q=cache:P72siWJTFjwJ:gato-docs.its.txstate.edu/slac/Subject/Math/Calculus/Findting-the-Equation-of-a-Tangent-Line/Finding%2520the%2520Equation%2520of%2520a%2520Tangent%2520Line.pdf+how+to+find+the+equation+of+a+tangent+line&hl=en&gl=us&sig=AFQjCNFcYzt1d-PU9hE2gbQKngp4FeRw3Q
Yes, the derivative of an equation is the slope of a line tangent to the graph.
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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.
By differentiating the answer and plugging in the x value along the curve, you are finding the exact slope of the curve at that point. In effect, this would be the slope of the tangent line, as a tangent line only intersects another at one point. To find the equation of a tangent line to a curve, use the point slope form (y-y1)=m(x-x1), m being the slope. Use the differential to find the slope and use the point on the curve to plug in for (x1, y1).
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).
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.
That line is [ y = 2 cos(2x) ].
a tangent is a line that touches the circle at only ONE point
For the equation (9x^2)/(x^2+4)