Calculus
Of One Real Variable – By Pheng Kim Ving 
2.1 
Return To Contents
Go To Problems & Solutions
1. Tangent Lines And Their Slopes 
NonVertical Tangent Lines
Thus, the equation of secant PQ
is y =
(6 + h)(x – 3) + 9. Similarly, T passes thru the point (3, 9) and has slope
6. Let
(x, y) be an arbitrary point of T. The equation of T is ( y – 9)/(x – 3) = 6, or y = 6(x – 3) + 9. In general, the equation
of the line passing thru the point (x_{0},
y_{0}) and having slope m is:
y = m(x
– x_{0}) + y_{0}.

Fig. 1.1 Tangent T is limit of secant PQ as Q approaches P. Note:
Scales on the axes are different. 
This equation is called the
pointslope equation of the line ( because it involves a point and the slope). Recall that the
slopeintercept equation of a line is y = mx + b ( because it involves the slope and the yintercept),
where m is the slope
and b the yintercept. We see that the equation y = mx + b is a special case of the equation y = m(x – x_{0}) + y_{0} where
x_{0} = 0 and thus y_{0} is the yintercept
and in the notation is replaced by the letter b.
Vertical Tangent Lines
The graph of y = x^{1/3} is illustrated in Fig. 1.2. As Q
approaches O, the secant OQ approaches the yaxis. The same
thing
occurs if Q approaches O from the part of the graph to the
left of O. Thus, the graph has a vertical line – the yaxis
in

Fig. 1.2 Tangent line to graph of y = x^{1/3} at x = 0 is yaxis, a vertical 
this case – as the tangent line at x = 0. The slope of that tangent line is:
then the graph of f
has a vertical tangent line at x = x_{0}.
The equation of such a vertical tangent line is x = x_{0}.
Note that horizontal tangent lines
are classified as nonvertical tangent lines. Their common slope is 0. The
equation of a
horizontal tangent line to the graph of y = f(x) at (x_{0}, y_{0}) is therefore y = y_{0}.
Where There Are No Tangent Lines
The graph of y = x
is sketched in Fig. 1.3. As Q_{1} approaches O, the line OQ_{1} stays the same,
as part of the graph to
the right of O. As Q_{2} approaches O, the line OQ_{2} stays the
same, as part of the graph to the left of O. So, if the graph

Fig. 1.3 Graph of y = x has no tangent line at x = 0. 
has a tangent line at x = 0, then there would be two distinct
tangent lines there, the right and left parts of the graph,
which contradicts the uniqueness of a tangent line. It follows that the graph
has no tangent line at x = 0. Now:
Definition 1.1
Suppose the function f(x) is continuous at x = x_{0}. Then the quotient:

If the tangent line to the graph of f(x) at x = a exists, then it's
clear that it must be unique. So we define its slope by
using the (twosided) limit, not onesided limits, which may be different when
they exist.
Definition 1.2
The slope of a curve C at a point P is the slope of the tangent line to C at P if such a tangent line exists. 
Example 1.1
a. Find the tangent line to the graph of y
= f(x) = x^{2} at x
= 3.
b. Find the tangent line to the
graph of f at x
= 0.
c. Use a graphing calculator or
software to sketch a graph of y = g(x) = x^{2/3}.
d. Does the graph of g
have a tangent line at x = 0?
Solution
a. When x = 3 we have y
= 3^{2} =
9. The slope of the tangent line at x = 3 is:
Thus the equation of the tangent line at x = 0 is y = 0(x – 0) + 0, or y = 0, which is the xaxis.
c. A graph of g is sketched in Fig. 1.4.
Fig. 1.4 y = x^{2/3}. 
d. The difference quotient of g at x = 0 is:
Hence the graph of g has no tangent line at x
= 0.
EOS
Go To Problems & Solutions Return To Top Of Page
2. Normal Lines 
Slopes Of Perpendicular Lines
Suppose two lines T and
N are perpendicular. If none of them is
vertical (thus none of them is horizontal), as shown in
Fig. 2.1, then their slopes are negative reciprocals of each other. In order to
see this, let the slope of T be m and that of

Fig. 2.1 Line N is normal to curve C at point P. 
N be n. Draw a vertical
line cutting T at A and N at B.
Angles PAH
and BPH are equal
because their sides are
perpendicular. So right triangles PAH and BPH are similar. It follows
that HA/PH = PH/HB. Now, m = HA/PH and n
= – HB/PH. Thus, m = –1/n, which is the same as n = –1/m.
Lines Normal To A Curve
A line is said to be normal to
a curve at a point if it's perpendicular to the tangent line of the curve at
that point. In Fig.
2.1, T is
tangent to curve C at point P, and N is normal to curve C at point P.
We have (slope
of normal N ) =
–1/(slope of
tangent T ).
Example 2.1
Find the equation of the normal line to the curve y = x^{2} at the point (1, 1).
Solution
Let f(x) = x^{2}. The slope of
the tangent line at the point (1, 1) is:
EOS
1. Find the equation of the tangent line to each
of the following curves at the indicated point.
Solution
b. Let f(x) =
ax^{2}
+
bx
+
c. The slope of the
tangent line is:
When x = u we have y =
au^{2}
+
bu
+
c. Thus the equation of
the tangent line is y = (2au + b)(x
– u) + (au^{2} + bu +
c), or y = (2au
+ b)x – au^{2} + c.
2. Find the equation of the normal line to each
of the following curves at the indicated point.
Solution
b. Let f(x) =
ax^{2}
+
bx
+
c. The slope of the
tangent line is:
Solution
a. dom(
g )
is the set of all nonnegative real numbers.
b. The points are (0, 0), (1, 1), (4, 2), and
(9, 3).
c. The slope of the tangent line is:
The slope of the
normal line is – 4, and thus its equation is y
= – 4(x – 4) + 2, or:
y = – 4x + 18.
d. The lines are drawn in the graph in part b.
4. Determine the
slope of the curve y =
x^{2}
– 1 at the
point
x
= a. Find
the equation of the tangent line with a slope of – 3
to that curve.
Solution
We'll show that the line y = – 3x – 13/4, which has a slope of –3, is
tangent to the curve y = x^{2}
– 1.
Let f(x) =
x^{2}
– 1. The
slope of f
at
x
=
a
is the same as the slope of the tangent line to f at
x
= a, so
it is:
5. Find all points on the curve y = x^{3} – 3x where the tangent line is parallel to the xaxis.
Solution
We'll show that the tangent lines
to the curve y = x^{3} – 3x
that are parallel the xaxis are at the points (1, –2) and
(–1, 2).
Let
g(x) = x^{3}
– 3x. The slope of a tangent line to
g
at an arbitrary point
x
is:
Return To Top Of Page Return To Contents