Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken
2
then they have the same slopes, so
f
0
(
a
) = 4
a
+ 3 = 6
.
Consequently,
a
=
3
4
.
keywords: tangent line, parallel, slope, deriva
tive
003
(part 1 of 1) 10 points
Find the
x
intercept of the tangent line at
the point
P
(

2
, f
(

2)) on the graph of
f
when
f
is defined by
f
(
x
) = 2
x
2

3
x
+ 2
.
1.
x
intercept =
6
11
2.
x
intercept =
11
6
3.
x
intercept =

6
11
correct
4.
x
intercept =

11
6
5.
x
intercept = 6
6.
x
intercept =

6
Explanation:
The slope,
m
, of the tangent line at the
point
P
(

2
, f
(

2)) on the graph of
f
is the
value of the derivative
f
0
(
x
) = 4
x

3
at
x
=

2,
i.e.
,
m
=

11.
On the other
hand,
f
(

2) = 16. Thus by the pointslope
formula an equation for the tangent line at
P
(

2
, f
(

2)) is
y

16 =

11(
x
+ 2)
,
i
.
e
., y
=

11
x

6
.
Consequently,
x
intercept =

6
11
.
keywords: tangent line, xintercept, slope
004
(part 1 of 1) 10 points
Find an equation for the tangent line to the
graph of
g
at the point
P
(1
, g
(1)) when
g
(
x
) = 5

x
3
.
1.
y
+ 3
x
+ 7 = 0
2.
y
= 7
x

3
3.
y
+ 7
x
+ 3 = 0
4.
y
= 3
x
+ 7
5.
y
+ 3
x
= 7
correct
Explanation:
If
x
= 1, then
g
(1) = 4. Thus the Newto
nian different quotient for
g
(
x
) = 5

x
3
at the point (1
,
4) becomes
g
(1 +
h
)

g
(1)
h
=
h
5

(1 +
h
)
3
i

4
h
=
5

h
3

3
h
2

3
h

5
h
.
Thus
g
0
(1) =
lim
h
→
0
(

h
2

3
h

3
)
=

3
.
Consequently, by the pointslope formula, an
equation for the tangent line to the graph of
g
at
P
is
y

4 =

3(
x

1)