Tangent Lines

TANGENT LINES

y = f(x)

There is a graphical interpretation of differentiability and of the number f '(a). Let us consider a generic function f and a point a in the domain of f. The graph of such a function is depicted in the box to the right. You will note that this graph is curved, but is fairly smooth, whatever that means.

What we would like for you to do now is to take a close look at this graph in the vicinity of the point (a,f(a)). Push the button to take a really close look.

The closer you look at the graph near (a,f(a)), the more the graph looks just like a straight line. We call this phenomenon "local linearity." We also call it "differentiability!"

Our goal on this page is to see that if f is differentiable at a, then the graph of f looks like a line near (a,f(a)). The line in question is the tangent line to the graph of f at the point (a,f(a)). The point (a,f(a)) is the point of tangency. Even better, the slope of this tangent line is f '(a). With all of this understood, we will be able to find equations of tangent lines.

**Existence of the Tangent Line at (a,f(a))**
If f is differentiable at a, then the graph of f has a nonvertical tangent line at (a,f(a)).

When f is differentiable at a, an equation that describes the tangent line can be obtained as follows.

**Equation of the Tangent Line at (a,f(a))**
The tangent line to y = f(x) at (a,f(a)) has these attributes: Point: (a,f(a)) Slope: f '(a) An equation for the tangent line arises from the point-slope equation for a line. y - f(a) = f '(a)(x - a)

Example. Let f(x) = sqrt(x), so that f '(x) = 1/2sqrt(x) for all positive real numbers x. The equation of the tangent line at a = 16 can be found from these pieces of data.

Point: (16,f(16)) = (16,sqrt(16)) = (16,4)
Slope: f '(16) = 1/2sqrt(16) = 1/8

An equation of the tangent line is thus the following: y - 4 = (x - 16)/8. Another perfectly good equation for the same line would be x - 8y = -16.

**Two tangent lines**

Moving to another point on the graph of the square root function, we can find an equation of the tangent line that corresponds to the domain value a = 100 as follows.

Point: (100,f(100)) = (100,sqrt(100)) = (100,10)
Slope: f '(100) = 1/2sqrt(100) = 1/20

An equation of the tangent line is thus the following: y - 10 = (x - 100)/20.

OK, nice pictures. The next task is to see WHY the derivative determines the slope of the tangent line. For this, we must consider the concept of a secant line.

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