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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 formula
http://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
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Derivative as the Slope of a Tangent?
Yes, the derivative of an equation is the slope of a line tangent to the graph.
What is an easy equation for finding the normal line?
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.
Find the slope of a tangent line to the graph?
Use the four-step process to find the slope of the tangent line to the graph of the given function at any point.
Find the equation of the circle if the circle is tangent to the line -x y 40 at the point 3-1 and the center is on the line x 2y-30?
x=9+11=5 11,2
Find the equation of the tangent line at the given value y equals ln x squared at x equals 1?
Given y=LN(x2). To find the equation of a tangent line to a point on a graph, we must figure out what the slope is at that exact point. We must take the derivitave with respect to x to come up with a function that will give the slope of the tangent line at any point on the graph. Any text book will tell you that the derivative with respect to X of LN(U) is the quantity U prime over U(In other words, the derivitive of U over U. U is whatever is inside the natural log). In this case, it is y = (2x/x2). To find the slope, we then plug in an x coordinate, 1. We get y=2. The slope of the tangent line at this point is 2. We now have m for the slope-intercept form y = mx + b. Now we must find b. B is the y intercept, aka where does the graph intersect the y axis. We use the slope. We know that the graph of the tangent line contains the point (0,1). We go down 2 and left one, to find that the tangent line intercepts the y axis at -1. We now have B. The equation of the tangent line of y = LN(x2) at X = 1 is y = 2X-1. Hope this helps!!!
