5 Steps to a 5 AP Calculus BC 2019

More Applications of Derivatives 177

Set 6x^2 = 6 ⇒x^2 =1orx=±1.

Atx=−1,y= 2 x^3 =2(−1)^3 =−2; (−1,−2) is a tangent point. Thus,y= 6 x+a⇒

− 2 =6(−1)+aora=4.

Atx=1,y= 2 x^3 =2(1)^3 =2; (1, 2) is a tangent point.

Thus,y= 6 x+a⇒ 2 =6(1)+aora=−4.

Therefore,a=±4.

Example 3

Find the coordinates of each point on the graph ofy^2 −x^2 − 6 x+ 7 =0 at which the tangent
line is vertical. Write an equation of each vertical tangent. (See Figure 9.1-6.)

− 701 x

y

y^2 − x^2 − 6 x + 7 = 0

x = –7 x = 1

Figure 9.1-6

Step 1: Find
dy
dx

.

y^2 −x^2 − 6 x+ 7 = 0

2 y

dy

dx

− 2 x− 6 = 0

dy

dx

=

2 x+ 6

2 y

=

x+ 3

y

Step 2: Find
dx
dy

.

Vertical tangent⇒

dx

dy

= 0.

dx

dy

=

1

dy/dx

=

1

(x+3)/y

=

y

x+ 3

Set

dx

dy

= 0 ⇒y= 0.