DEVELOPMENT OF THE LESSON
(A) INTRODUCTION
In the previous lesson, we defined the tangent line at a point P as the limiting position of the secant lines P Q, where Q is another point on the curve, as Q approaches P . There is a slight problem with this definition because we have no means of computing the limit of lines. Hence, we need to work on the numbers that characterize the lines.
Teaching Tip
Ask the class what things determine lines; in particular, what is the minimum that you should know so that you can draw one and only one line. Expect correct answers such as two points, slope and y-intercept, the two intercepts, a point and the slope.
Give prominence to the last of the list, emphasizing that there are infinitely many lines passing thru a point; there are infinitely many (parallel) lines with the same slope, but there is only one line passing through a point with a given slope.
• Recall the slope of a line passing through two points (x0, y0) and (x, y).
Recall: Slope of a Line
A line ` passing through distinct points (x0, y0) and (x, y) has slope m` = y y0
x x0
.
EXAMPLE 1: Given A(1, 3), B(3, 2), and C( 1, 0), what are the slopes of the lines AB, AC and BC?
Solution. The slope of AB is
mAB = 2 ( 3)
3 1 = 1
2. The slope of AC is
mAC = 0 ( 3)
1 1 = 3
2. The slope of BC is
mBC = 0 ( 2)
1 3 = 2
4 = 1 2.
.
• Recall the point-slope form of the equation of the line with slope m and passing through the point P (x0, y0).
Recall: Point-Slope Form
The line passing through (x0, y0)with slope m has the equation y y0 = m(x x0).
EXAMPLE 2: From Example 1 above, since mAB = 12, then using A(1, 3) as our point, then the point-slope form of the equation of AB is
y ( 3) = 1
2(x 1) or y + 3 = 1
2(x 1).
Teaching Tip
Ask the class what would happen if we had chosen B(3, 2) as our point instead of A. Answer: We would get an equivalent equation.
(B) LESSON PROPER
THE EQUATION OF THE TANGENT LINE
Given a function y = f(x), how do we find the equation of the tangent line at a point P (x0, y0)?
Consider the graph of a function y = f(x) whose graph is given below. Let P (x0, y0) be a point on the graph of y = f(x). Our objective is to find the equation of the tangent line (T L) to the graph at the point P (x0, y0).
y = f (x) y0 P
Q
x y
T L
x0
• Find any point Q(x, y) on the curve.
• Get the slope of this secant line P Q.
mP Q= y y0 .
• Observe that letting Q approach P is equivalent to letting x approach x0. Teaching Tip
Illustrate this by choosing points x1and x2in between x and x0and projecting it vertically to the corresponding points Q1 and Q2 on the graph. The class should be able to see this equivalence.
We use the formal definition of the tangent line:
• Since the tangent line is the limiting position of the secant lines as Q approaches P , it follows that the slope of the tangent line (T L) at the point P is the limit of the slopes of the secant lines P Q as x approaches x0. In symbols,
mT L= lim
x!x0
y y0
x x0 = lim
x!x0
f (x) f (x0) x x0 .
• Finally, since the tangent line passes through P (x0, y0), then its equation is given by y y0 = mT L(x x0).
Teaching Tip
It is up to you if you want your students to put this into standard (slope-intercept) form. This is of course not an objective but it helps for easy checking of final answers.
SUMMARY AND EXAMPLES
Equation of the Tangent Line
To find the equation of the tangent line to the graph of y = f(x) at the point P (x0, y0), follow this 2-step process:
• Get the slope of the tangent line by computing m = lim
x!x0
y y0
x x0 or m = lim
x!x0
f (x) f (x0) x x0 .
• Substitute this value of m and the coordinates of the known point P (x0, y0) into the point-slope form to get
y y0 = m(x x0).
EXAMPLE 3: Find the equation of the tangent line to y = x2 at x = 1.
Solution. To get the equation of the line, we need the point P (x0, y0) and the slope m. We are only given x0 = 2. However, the y-coordinate of x0 is easy to find by substituting x0= 2into y = x2. This gives us y0 = 4. Hence, P has the coordinates (2, 4).
Now, we look for the slope:
xlim!x0
y y0
x x0 = lim
x!2
x2 4 x 2 = 4.
Finally, the equation of the tangent line with slope m = 4and passing through P (2, 4) is
y 4 = 4(x 2) or y = 4x 4.
.
2 P y = x2
Teaching Tip
If you want to make your own examples, make sure that P is a point on the curve!
For example, it does not make sense to find the equation of the tangent line to y = x2 at the point P (2, 3) since P is not on the parabola (3 6= 22). You could modify this to P (2, 4) or better yet, for later examples, you can just ask for the tangent line at a specific x-coordinate.
EXAMPLE 4: Find the slope-intercept form of the tangent line to f(x) =p
xat x = 4.
4
P y =p
x
Solution. Again, we find the y-coordinate of x0 = 4: y0 = f (x0) = px0=p
4 = 2. Hence,
have to rationalize the numerator to evaluate the limit. The next example shows that our process of finding the tangent line works even for hori-zontal lines.
EXAMPLE 5: Show that the tangent line to y = 3x2 12x + 1 at the point (2, 11) is horizontal.
Solution. Recall that a horizontal line has zero slope. Now, computing for the slope, we get:
Since the slope of the tangent line is 0, it must be horizontal. Its equation is y ( 11) = 0(x 2) or y = 11.
Therefore, substituting this into the point-slope form with P (1, 5) and m = 2, we get y 5 = 2(x 1) i.e., y = 2x + 3.
This is the same equation as that of the given line. .
(C) EXERCISES
Find the standard (slope-intercept form) equation of the tangent line to the following functions at the specified points:
1. f(x) = 3x2 12x + 1at the point (0, 1) Answer: y = 12x + 1
It is not productive to dwell on finding the slope of the tangent line to very com-plicated functions. A more efficient way of finding this will be given in the next sections when we discuss derivatives.
(D) ENRICHMENT
This section explores the equation of vertical tangent lines. In the last topic, we remarked that vertical tangent lines may exist. However, we know that the slope of a vertical line does not exist or is undefined. How do we reconcile these seemingly contradicting ideas?
For example, consider the vertical tangent line to the graph of y = p3
Observe that the last expression is undefined
✓1 0
◆
if we substitute x with 0. Hence, the slope of this tangent line is undefined.
Therefore, our computation for the slope of the tangent line to this curve is actually con-sistent with our idea of the slope of a vertical line. The next question to ask is: Does this tangent line have an equation? The answer is yes. Recall that a vertical line passing through the point (x0, y0) possesses the equation x = x0.
Equation of Vertical Tangent Lines
Let f be a function that is continuous at x0. Assuming that the tangent line to the graph of y = f(x) at the point P (x0, y0) is vertical, then its equation is
x = x0.