Tangent Lines

Section 2.1

Secant Line

A secant line is a line that connects two points on a graph.

1)( 2 xxfNotice the slopes of secant lines are different depending which two points you connect.

Secant Line

1)( 2 xxf

Up to this point we have used the formula

12

12

xx

yym

to find the slope of the secant line joining points (x1, y1) and

(x2, y2).

This formula has four inputs x1 and x2,

y1, y2.. Once we input these four

values into the formula, our output represents the slope of that secant line.

Slope of Secant LineWe will now find another formula for the slope of the secant line between two points.

We will rename the points (x1, y1) and (x2, y2) using function notation in order to obtain our new slope formula.

(x1, y1)

(x2, y2)

We will let x1 = x. Then it follows that y1, the y-value for x1, can be rewritten in

function notation as f(x). Note that f(x) refers to the output when the input is x.

Similarly, we rename x2 as x + h, where h refers to the distance from x1 to x2.

Then using function notation, y2 will be rewritten as f(x + h), the output when the input is x + h.

Point (x1, y1) becomes (x, f(x))andPoint (x2, y2) becomes (x + h, f(x + h))

12

12

xx

yym

The slope formula

2y 1y

h

xfhxf

xhx

xfhxf

xhx

xfhxfm

)()(

)()(

)(

)()(

x1

Simplify the Denominator

x2

Recall that h is the distance from x1 to x2, namely x2 – x1

becomes

Note: This formula for the slope has only two inputs: x, the smallest x-value and h, where h is the distance from the first x-value to the second x-value.

Example: Find the formula for the slope of any secant line for the function22)( xxf

.

h

xfhxfm

)()(sec

Step 1: Find :)( hxf 2)(2)( hxhxf 22 242 hxhx

h

xfhxf )()(h

xhxhx )2()242( 222

Step 2: Substitute into the formula:

hx 24 This is a function whose output is slope of a secant line forinputs x (the first x coordinate) and h (the distance between the x’s).

For example:

Find the slope of the secant line to the graph of from x = -1 to x = 2.

22)( xxf

We know the slope of the secant lines of the function f(x) =-2x2 follow the formula: hx 24

In this case, x = -1 (first x value from left to right) and h = 3 (distance from x = -1 to x = 2).

Therefore, the slope of the secant line shown is = -4(-1) – 2(3) = 4 – 6 = -2

Slope = -2

Tangent LineTangent Line: The tangent line is a line drawn at a single point on a graph.

How do you draw a tangent line at an x-value?Think of having a rock at the end of a string and following some curve with this rock.

If you release the string at a point, say at x = ½, the path the rock follows is your tangent line at x = ½ .

Likewise, the path the rock follows if released at x = 3 would be the tangent line drawn at x = 3.

A tangent line can be drawn at each point.

Finding the slope of the tangent line.

Notice we cannot use the

formula

because it would require two points on the line. We only know one point, the point of tangency (-1, -4)

12

12

xx

yym

Tangent line at x = -1.How do we find the slope of the

tangent line, say at x = -1 for the graph of f(x) = x3 – 3x2?

Tangent line at x = -1.

We cannot use

because wedo not have h, the distance b/w the two x-values.

h

xfhxf )()(

Let’s draw a secant line from x = -1 to x = any other x- value.

Obviously, the slope of this secant line is different from the slope of our tangent line.

Secant line.

Let’s allow the second x-value get closer to x = -1 and draw the secant line again.

Again let’s allow the second x-value get closer to x = -1 and draw the secant line.

Notice that although the secant line is different from the tangent line, they are getting closer together as the second x-value gets closer to x = -1.

Again let’s allow the second x-value get closer to x = -1 and draw the secant line.

If we continue to allow the second x-value closer to x = -1 then the secant line will approach the tangent line.

h

xfhxf )()( Notice that the second point approaching x = -1 simply means that the distance b/w the two x-values is approaching 0. (h 0)

Therefore, to find the slope of the tangent line…

we find the slope of the

secant line,

then take the limit as h0

h

xfhxf )()(

linesec0

linetan limmmh

Find the slope of the tangent line to f(x) = x3 – 3x2 at x = -1.

Example-Polynomial

Given the equation 2y x 4x, 4, 0 a) Find the slope of a secant line through the given point

2 2

sec

x h 4 x 4xx

h

hm

2

sec

2 2x 2xh h 4x x 4xm

h

4h 2

sec

2xh h 4hm

h

sec

h 2m

h

x h 4

sec 2 hm x 4

Given the equation 2y x 4x, 4, 0 b) Find the SLOPE of a tangent line through the given point

From part a, sec 2 hm x 4

tanh 0

2x hm lim 2x 44

At (4, 0), 2x – 4 = 2(4) – 4 = 4

c) Find the EQUATION of a tangent line through the given point

From part b, we have the slope (4)….and we have the pt (4, 0)

y – 0 = 4(x – 4)

c) Find the value of x for which the slope of the tangent line is 0

2x – 4 = 0x = 2

Given the equation 3y x 2x, 1, 1 a) Find the slope of a secant line through the given point

3 3

sec

x h 2 x 2xx

h

hm

3 2 2 3 3

sec

x 3x h 3xh h 2x x 2xm

h

2h 2

sec

2 33x h 3xh h 2hm

h

2 2

sec

h 3x 3x 2m

h

h

h 2

sec23x hm 3xh 2

b) Find the SLOPE of a tangent line through the given point

From part a,

tanh

2 22

03x 3xhm l m 2i 3h x 2

At (1, -1),

c) Find the EQUATION of a tangent line through the given point

From part b, we have the slope (1)….and we have the pt (1, -1)

y + 1 = 1(x - 1)

c) Find the value of x for which the slope of the tangent line is 0

Given the equation 3y x 2x, 1, 1

2sec

23x hm 3xh 2

223x 2 3 1 2 1

2 2 2 23x 2 0 x x

3 3

Using the limit definition, find the first derivative of 3f x x

3 3

h 0

x xl

h

him

3 33

h

2

0

2x 3x h 3xh xlim

h

h

2 2 3

h 0

3x h 3xh hlim

h

0

2 2

h

h 3x 3xh hlim

h

2 2 2

h 03x 3xhi xl h 3m

Using the limit definition, find the first derivative of f x 1 x

h 0

1 1 xlim

h

x h

h 0

xlim

h

x h

x

x x

h x

h

h 0

xlim

h x h

h

x

x

h 0

hlim

h x h x

h 0

1 1lim

2 xx h x

Using the limit definition, find the first derivative of f x x 1

h 0

x h 1 x 1lim

h

h 0

x h x1 x 1 1 x 1lim

h h xx 1

h

1

h 0

1 x 1lim

h 1

h

x 1

x

x h

h 0

h

x hlim

h 1 x 1

h 0

1 1lim

2 x 11 xx h 1