Math - slope of tangent line?

Question

How do I find the slope of the tangent line to the graph of y= (x+1)/(x-3) passing through the point (1,5)??
Do I use derivatives? Or is there any other way? How would I use derivatives?

Answer 1

Yes, to use derivative ONLY when the given point is on the curve. In that case, dy/dx at the given point is the slope of the tangent line. Once you get the slope, you can get the line with the point-slope form easily.
Now, in this case the given point is NOT on the curve. You do not know the tangent point thus you do not know the slope. Of course you may still write the equation:
y(x) - 5 = (dy/dx) (x - 1) and solve it (where dy/dx is in terms of x, thus the equation above only contains x). But here we may solve the problem with a different method.

Let the tangent line equation to be: y - 5 = k(x - 1)
since it passes through Point (1,5). This line must have cross-section with the given curve. Putting y = k(x - 1) + 5 into y= (x+1)/(x-3), we get
k(x - 1) + 5 = (x + 1) /(x - 3), a quadratic equation in x with k as a parameter. Since there is ONLY one tangent point thus this equation must have a single solution. That means the discriminent (which involves k only) must be zero.
Now, it is your task to find the discriminent of the quadratic equation, and set it to be zero and thus enable you to get k, and finally you get the tangent line y = k(x - 1) + 5

Answer 2

Calculus
The concept of a slope is central to differential calculus. For non-linear functions, the rate of change varies along the curve. The derivative of the function at a point is the slope of the line tangent to the curve at the point, and is thus equal to the rate of change of the function at that point.

If we let Δx and Δy be the distances (along the x and y axes, respectively) between two points on a curve, then the slope given by the above definition,

,
is the slope of a secant line to the curve. For a line, the secant between any two points is the line itself, but this is not the case for any other type of curve.

For example, the slope of the secant intersecting y = x² at (0,0) and (3,9) is m = (9 - 0) / (3 - 0) = 3 (which happens to be the slope of the tangent at, and only at, x = 1.5, a consequence of the mean value theorem).

By moving the two points closer together so that Δy and Δx decrease, the secant line more closely approximates a tangent line to the curve, and as such the slope of the secant approaches that of the tangent. Using differential calculus, we can determine the limit, or the value that Δy/Δx approaches as Δy and Δx get closer to zero; it follows that this limit is the exact slope of the tangent. If y is dependent on x, then it is sufficient to take the limit where only Δx approaches zero. Therefore, the slope of the tangent is the limit of Δy/Δx as Δx approaches zero. We call this limit the derivative.


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